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Why does the vacuum even have permeability and permittivity?
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up vote
4
down vote
favorite
The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why?
Why are the vacuum permeability and permittivity non-zero and non-infinite?
What would a universe in which the vacuum permeability and permittivity were zero look like?
What would a universe in which the vacuum permeability and permittivity were infinite look like?
electromagnetism vacuum
asked 2 hours ago
cowlinator
1825
add a comment |
up vote
4
down vote
favorite
The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why?
Why are the vacuum permeability and permittivity non-zero and non-infinite?
What would a universe in which the vacuum permeability and permittivity were zero look like?
What would a universe in which the vacuum permeability and permittivity were infinite look like?
electromagnetism vacuum
asked 2 hours ago
cowlinator
1825
2
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
1
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why?
Why are the vacuum permeability and permittivity non-zero and non-infinite?
What would a universe in which the vacuum permeability and permittivity were zero look like?
What would a universe in which the vacuum permeability and permittivity were infinite look like?
electromagnetism vacuum
asked 2 hours ago
cowlinator
1825
The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why?
Why are the vacuum permeability and permittivity non-zero and non-infinite?
What would a universe in which the vacuum permeability and permittivity were zero look like?
What would a universe in which the vacuum permeability and permittivity were infinite look like?
electromagnetism vacuum
electromagnetism vacuum
asked 2 hours ago
cowlinator
1825
asked 2 hours ago
cowlinator
1825
asked 2 hours ago
cowlinator
1825
asked 2 hours ago
cowlinator
1825
asked 2 hours ago
cowlinator
1825
1825
2
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
1
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago
add a comment |
2
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
1
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago
2
2
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
1
1
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
I will assume throughout this answer that we fix the value of $c$ independently of $epsilon_0$ or $mu_0$. The vacuum permittivity and permeability are related to one another by $varepsilon_0mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electric and magnetism are both manifestations of the same fundamental force.
The permittivity is related to the dimensionless fine structure constant $alpha$ by $alpha = frac14pivarepsilon_0 frace^2hbar c$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $varepsilon_0$.
If we take $alpha rightarrow 0$ ($varepsilon_0 rightarrow infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $alpha rightarrow infty$ ($varepsilon_0 rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
beginequation
F = frac14pivarepsilon_0 fracq q'r^2,
endequation
where the former limit gives $Frightarrow 0$ and the latter $F rightarrow infty$ for finite charges and distances.
answered 2 hours ago
d_b
938614
add a comment |
up vote
2
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if they were zero, you recover Newtonian mechanics (the speed of light would become infinite), if it is infinite, there is no recognizable universe, you would have frozen light (zero speed) and thus neither mass nor reference frames could exist
answered 2 hours ago
Wolphram jonny
10.4k22451
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
add a comment |
up vote
1
down vote
Why are the vacuum permeability and permittivity non-zero and non-infinite?
Why not?
What would a universe in which the vacuum permeability and permittivity were zero look like?
Because $$c = frac1sqrtepsilon_0 mu_0,$$
you'd have an infinite speed of light as Wolphram jonny pointed out.
What would a universe in which the vacuum permeability and permittivity were infinite look like?
Wolphram jonny answered this and it follows from the equation I gave.
answered 2 hours ago
PiKindOfGuy
574514
add a comment |
up vote
0
down vote
Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
answered 1 hour ago
G. Smith
1,10718
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I will assume throughout this answer that we fix the value of $c$ independently of $epsilon_0$ or $mu_0$. The vacuum permittivity and permeability are related to one another by $varepsilon_0mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electric and magnetism are both manifestations of the same fundamental force.
The permittivity is related to the dimensionless fine structure constant $alpha$ by $alpha = frac14pivarepsilon_0 frace^2hbar c$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $varepsilon_0$.
If we take $alpha rightarrow 0$ ($varepsilon_0 rightarrow infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $alpha rightarrow infty$ ($varepsilon_0 rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
beginequation
F = frac14pivarepsilon_0 fracq q'r^2,
endequation
where the former limit gives $Frightarrow 0$ and the latter $F rightarrow infty$ for finite charges and distances.
answered 2 hours ago
d_b
938614
add a comment |
up vote
3
down vote
accepted
I will assume throughout this answer that we fix the value of $c$ independently of $epsilon_0$ or $mu_0$. The vacuum permittivity and permeability are related to one another by $varepsilon_0mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electric and magnetism are both manifestations of the same fundamental force.
The permittivity is related to the dimensionless fine structure constant $alpha$ by $alpha = frac14pivarepsilon_0 frace^2hbar c$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $varepsilon_0$.
If we take $alpha rightarrow 0$ ($varepsilon_0 rightarrow infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $alpha rightarrow infty$ ($varepsilon_0 rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
beginequation
F = frac14pivarepsilon_0 fracq q'r^2,
endequation
where the former limit gives $Frightarrow 0$ and the latter $F rightarrow infty$ for finite charges and distances.
answered 2 hours ago
d_b
938614
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I will assume throughout this answer that we fix the value of $c$ independently of $epsilon_0$ or $mu_0$. The vacuum permittivity and permeability are related to one another by $varepsilon_0mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electric and magnetism are both manifestations of the same fundamental force.
The permittivity is related to the dimensionless fine structure constant $alpha$ by $alpha = frac14pivarepsilon_0 frace^2hbar c$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $varepsilon_0$.
If we take $alpha rightarrow 0$ ($varepsilon_0 rightarrow infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $alpha rightarrow infty$ ($varepsilon_0 rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
beginequation
F = frac14pivarepsilon_0 fracq q'r^2,
endequation
where the former limit gives $Frightarrow 0$ and the latter $F rightarrow infty$ for finite charges and distances.
answered 2 hours ago
d_b
938614
I will assume throughout this answer that we fix the value of $c$ independently of $epsilon_0$ or $mu_0$. The vacuum permittivity and permeability are related to one another by $varepsilon_0mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electric and magnetism are both manifestations of the same fundamental force.
The permittivity is related to the dimensionless fine structure constant $alpha$ by $alpha = frac14pivarepsilon_0 frace^2hbar c$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $varepsilon_0$.
If we take $alpha rightarrow 0$ ($varepsilon_0 rightarrow infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $alpha rightarrow infty$ ($varepsilon_0 rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
beginequation
F = frac14pivarepsilon_0 fracq q'r^2,
endequation
where the former limit gives $Frightarrow 0$ and the latter $F rightarrow infty$ for finite charges and distances.
answered 2 hours ago
d_b
938614
answered 2 hours ago
d_b
938614
answered 2 hours ago
d_b
938614
answered 2 hours ago
d_b
938614
938614
add a comment |
add a comment |
up vote
2
down vote
if they were zero, you recover Newtonian mechanics (the speed of light would become infinite), if it is infinite, there is no recognizable universe, you would have frozen light (zero speed) and thus neither mass nor reference frames could exist
answered 2 hours ago
Wolphram jonny
10.4k22451
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
add a comment |
up vote
2
down vote
if they were zero, you recover Newtonian mechanics (the speed of light would become infinite), if it is infinite, there is no recognizable universe, you would have frozen light (zero speed) and thus neither mass nor reference frames could exist
answered 2 hours ago
Wolphram jonny
10.4k22451
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
add a comment |
up vote
2
down vote
up vote
2
down vote
if they were zero, you recover Newtonian mechanics (the speed of light would become infinite), if it is infinite, there is no recognizable universe, you would have frozen light (zero speed) and thus neither mass nor reference frames could exist
answered 2 hours ago
Wolphram jonny
10.4k22451
if they were zero, you recover Newtonian mechanics (the speed of light would become infinite), if it is infinite, there is no recognizable universe, you would have frozen light (zero speed) and thus neither mass nor reference frames could exist
answered 2 hours ago
Wolphram jonny
10.4k22451
answered 2 hours ago
Wolphram jonny
10.4k22451
answered 2 hours ago
Wolphram jonny
10.4k22451
answered 2 hours ago
Wolphram jonny
10.4k22451
10.4k22451
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
add a comment |
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
Could you explain how infinite speed of light means light is frozen?
– Schwern
18 mins ago
add a comment |
up vote
1
down vote
Why are the vacuum permeability and permittivity non-zero and non-infinite?
Why not?
What would a universe in which the vacuum permeability and permittivity were zero look like?
Because $$c = frac1sqrtepsilon_0 mu_0,$$
you'd have an infinite speed of light as Wolphram jonny pointed out.
What would a universe in which the vacuum permeability and permittivity were infinite look like?
Wolphram jonny answered this and it follows from the equation I gave.
answered 2 hours ago
PiKindOfGuy
574514
add a comment |
up vote
1
down vote
Why are the vacuum permeability and permittivity non-zero and non-infinite?
Why not?
What would a universe in which the vacuum permeability and permittivity were zero look like?
Because $$c = frac1sqrtepsilon_0 mu_0,$$
you'd have an infinite speed of light as Wolphram jonny pointed out.
What would a universe in which the vacuum permeability and permittivity were infinite look like?
Wolphram jonny answered this and it follows from the equation I gave.
answered 2 hours ago
PiKindOfGuy
574514
add a comment |
up vote
1
down vote
up vote
1
down vote
Why are the vacuum permeability and permittivity non-zero and non-infinite?
Why not?
What would a universe in which the vacuum permeability and permittivity were zero look like?
Because $$c = frac1sqrtepsilon_0 mu_0,$$
you'd have an infinite speed of light as Wolphram jonny pointed out.
What would a universe in which the vacuum permeability and permittivity were infinite look like?
Wolphram jonny answered this and it follows from the equation I gave.
answered 2 hours ago
PiKindOfGuy
574514
Why are the vacuum permeability and permittivity non-zero and non-infinite?
Why not?
What would a universe in which the vacuum permeability and permittivity were zero look like?
Because $$c = frac1sqrtepsilon_0 mu_0,$$
you'd have an infinite speed of light as Wolphram jonny pointed out.
What would a universe in which the vacuum permeability and permittivity were infinite look like?
Wolphram jonny answered this and it follows from the equation I gave.
answered 2 hours ago
PiKindOfGuy
574514
answered 2 hours ago
PiKindOfGuy
574514
answered 2 hours ago
PiKindOfGuy
574514
answered 2 hours ago
PiKindOfGuy
574514
574514
add a comment |
add a comment |
up vote
0
down vote
Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
answered 1 hour ago
G. Smith
1,10718
add a comment |
up vote
0
down vote
Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
answered 1 hour ago
G. Smith
1,10718
add a comment |
up vote
0
down vote
up vote
0
down vote
Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
answered 1 hour ago
G. Smith
1,10718
Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
answered 1 hour ago
G. Smith
1,10718
edited 52 mins ago
answered 1 hour ago
G. Smith
1,10718
answered 1 hour ago
G. Smith
1,10718
answered 1 hour ago
G. Smith
1,10718
1,10718
add a comment |
add a comment |
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2
Any zero or infinite value would imply that the wave propagating in vacuum would have the electric and magnetic intensities independent of each other but then neither the Ampere-Maxwell nor the Faraday law would make any sense. When finite the actual ratio ie., the vacuum impedance is just a number representing the choice of units.
– hyportnex
1 hour ago
1
Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0
– Pieter
1 hour ago