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Does an electromagnetic field affect neutral particles via the metric because of the EM stress-energy tensor?
Clash Royale CLAN TAG#URR8PPP
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I'm just starting to learn general relativity (GR), and I'm a beginner, but I came out with this situation which is unclear to me: The trajectory of a charged particle in GR is given from the equation:
$$dotu^mu + Gamma^mu_alpha beta u^alpha u^beta = fracqm F^mu_; nu , u^nu$$
So, if I have a neutral particle $q=0$ the equation reduces to the geodesic equation for a free particle, but because of the Einstein-Maxwell equations:
$$R_mu nu - frac12 R g_mu nu = T^EM_mu nu$$
the EM stress-energy tensor determines the form of the metric, and consequently the Christoffel symbols that appears in the geodesic equation for the neutral particle. So would the trajectory of this neutral particle in an EM field be different from the case of a space-time with a null EM field?
electromagnetism general-relativity stress-energy-momentum-tensor geodesics
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
asked Aug 19 at 0:28
Julian Ar.
3291210
add a comment |Â
up vote
12
down vote
favorite
I'm just starting to learn general relativity (GR), and I'm a beginner, but I came out with this situation which is unclear to me: The trajectory of a charged particle in GR is given from the equation:
$$dotu^mu + Gamma^mu_alpha beta u^alpha u^beta = fracqm F^mu_; nu , u^nu$$
So, if I have a neutral particle $q=0$ the equation reduces to the geodesic equation for a free particle, but because of the Einstein-Maxwell equations:
$$R_mu nu - frac12 R g_mu nu = T^EM_mu nu$$
the EM stress-energy tensor determines the form of the metric, and consequently the Christoffel symbols that appears in the geodesic equation for the neutral particle. So would the trajectory of this neutral particle in an EM field be different from the case of a space-time with a null EM field?
electromagnetism general-relativity stress-energy-momentum-tensor geodesics
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
asked Aug 19 at 0:28
Julian Ar.
3291210
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
I'm just starting to learn general relativity (GR), and I'm a beginner, but I came out with this situation which is unclear to me: The trajectory of a charged particle in GR is given from the equation:
$$dotu^mu + Gamma^mu_alpha beta u^alpha u^beta = fracqm F^mu_; nu , u^nu$$
So, if I have a neutral particle $q=0$ the equation reduces to the geodesic equation for a free particle, but because of the Einstein-Maxwell equations:
$$R_mu nu - frac12 R g_mu nu = T^EM_mu nu$$
the EM stress-energy tensor determines the form of the metric, and consequently the Christoffel symbols that appears in the geodesic equation for the neutral particle. So would the trajectory of this neutral particle in an EM field be different from the case of a space-time with a null EM field?
electromagnetism general-relativity stress-energy-momentum-tensor geodesics
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
asked Aug 19 at 0:28
Julian Ar.
3291210
I'm just starting to learn general relativity (GR), and I'm a beginner, but I came out with this situation which is unclear to me: The trajectory of a charged particle in GR is given from the equation:
$$dotu^mu + Gamma^mu_alpha beta u^alpha u^beta = fracqm F^mu_; nu , u^nu$$
So, if I have a neutral particle $q=0$ the equation reduces to the geodesic equation for a free particle, but because of the Einstein-Maxwell equations:
$$R_mu nu - frac12 R g_mu nu = T^EM_mu nu$$
the EM stress-energy tensor determines the form of the metric, and consequently the Christoffel symbols that appears in the geodesic equation for the neutral particle. So would the trajectory of this neutral particle in an EM field be different from the case of a space-time with a null EM field?
electromagnetism general-relativity stress-energy-momentum-tensor geodesics
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
asked Aug 19 at 0:28
Julian Ar.
3291210
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
edited Aug 19 at 9:40
Peter Mortensen
1,87811223
1,87811223
asked Aug 19 at 0:28
Julian Ar.
3291210
asked Aug 19 at 0:28
Julian Ar.
3291210
asked Aug 19 at 0:28
Julian Ar.
3291210
3291210
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
11
down vote
accepted
The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).
The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.
Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.
edited Aug 19 at 10:46
Jens
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
add a comment |Â
up vote
4
down vote
Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.
answered Aug 19 at 3:28
knzhou
33.4k897167
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).
The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.
Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.
edited Aug 19 at 10:46
Jens
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
add a comment |Â
up vote
11
down vote
accepted
The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).
The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.
Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.
edited Aug 19 at 10:46
Jens
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).
The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.
Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.
edited Aug 19 at 10:46
Jens
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).
The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.
Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.
edited Aug 19 at 10:46
Jens
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
edited Aug 19 at 10:46
Jens
2,38611431
edited Aug 19 at 10:46
Jens
2,38611431
edited Aug 19 at 10:46
Jens
2,38611431
2,38611431
answered Aug 19 at 2:21
Someone
1,16611029
answered Aug 19 at 2:21
Someone
1,16611029
answered Aug 19 at 2:21
Someone
1,16611029
1,16611029
add a comment |Â
add a comment |Â
up vote
4
down vote
Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.
answered Aug 19 at 3:28
knzhou
33.4k897167
add a comment |Â
up vote
4
down vote
Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.
answered Aug 19 at 3:28
knzhou
33.4k897167
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.
answered Aug 19 at 3:28
knzhou
33.4k897167
Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.
answered Aug 19 at 3:28
knzhou
33.4k897167
answered Aug 19 at 3:28
knzhou
33.4k897167
answered Aug 19 at 3:28
knzhou
33.4k897167
answered Aug 19 at 3:28
knzhou
33.4k897167
33.4k897167
add a comment |Â
add a comment |Â
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