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Is light actually faster than what our present measurements tell us?
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It is well established that the light speed in a perfect vacuum is roughly $3times 10^8 :rm m/s$. But it is also known that outer space is not a perfect vacuum, but a hard vacuum. So, is the speed limit theoretically faster than what we can measure empirically, because the hard vacuum slows it down? Is this considered when measuring distances with light?
special-relativity cosmology experimental-physics speed-of-light refraction
asked Sep 5 at 15:01
William
3371212
 |Â
show 12 more comments
up vote
38
down vote
favorite
It is well established that the light speed in a perfect vacuum is roughly $3times 10^8 :rm m/s$. But it is also known that outer space is not a perfect vacuum, but a hard vacuum. So, is the speed limit theoretically faster than what we can measure empirically, because the hard vacuum slows it down? Is this considered when measuring distances with light?
special-relativity cosmology experimental-physics speed-of-light refraction
asked Sep 5 at 15:01
William
3371212
8
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
8
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
36
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
8
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
11
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03
 |Â
show 12 more comments
up vote
38
down vote
favorite
up vote
38
down vote
favorite
It is well established that the light speed in a perfect vacuum is roughly $3times 10^8 :rm m/s$. But it is also known that outer space is not a perfect vacuum, but a hard vacuum. So, is the speed limit theoretically faster than what we can measure empirically, because the hard vacuum slows it down? Is this considered when measuring distances with light?
special-relativity cosmology experimental-physics speed-of-light refraction
asked Sep 5 at 15:01
William
3371212
It is well established that the light speed in a perfect vacuum is roughly $3times 10^8 :rm m/s$. But it is also known that outer space is not a perfect vacuum, but a hard vacuum. So, is the speed limit theoretically faster than what we can measure empirically, because the hard vacuum slows it down? Is this considered when measuring distances with light?
special-relativity cosmology experimental-physics speed-of-light refraction
asked Sep 5 at 15:01
William
3371212
edited Sep 6 at 0:27
asked Sep 5 at 15:01
William
3371212
asked Sep 5 at 15:01
William
3371212
asked Sep 5 at 15:01
William
3371212
3371212
8
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
8
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
36
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
8
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
11
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03
 |Â
show 12 more comments
8
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
8
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
36
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
8
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
11
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03
8
8
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
8
8
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
36
36
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
8
8
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
11
11
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03
 |Â
show 12 more comments
5 Answers
5
active
oldest
votes
up vote
95
down vote
accepted
If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $Delta c/c$ of $3 times 10^-4$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ fracDelta cc = 3 times 10^-4 , P $$
In high vacuum labs we can, without too much effort, get to $10^-10$ torr and this is around $10^-13$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ fracDelta cc approx 3 times 10^-17 $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
answered Sep 5 at 15:55
John Rennie
263k41512761
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
 |Â
show 2 more comments
up vote
70
down vote
The answer by John Rennie is good so far as the impacts of the imperfect vacuum go, so I won't repeat that here.
As regards the last part of your question about whether this should be accounted when measuring distance, it's worth noting that the standard defines the speed of light to be a specific value and then, using also the definition of the second, derives the meter as a matter of measurement. So as the standards are currently written, the speed of light is exact by definition.
Your question, as written, implicitly assumes that the meter and the second are given by definition and the speed of light a question of measurement.
So from that perspective, your question really should be written to ask whether the impact of imperfect vacuum impacts our definition of the meter. The answer to that, is that it probably does, as was approximately quantified by John Rennie. Whether or not it is important depends on what method is used and what other experimental uncertainties are inherent to that method.
answered Sep 5 at 16:48
Brick
1,194414
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
add a comment |Â
up vote
19
down vote
There is a constant in physics called $c$ that is the "exchange rate" between space and time. One second in time is in some sense "equivalent" to $c$ times one second (which then gives a distance in space). Light is taken to travel at $c$. Note that $c$ isn't the speed of light, but rather the speed of light is $c$, which is a subtle distinction ($c$ being what it is causes light to travel at that speed, rather than light traveling at that speed causes $c$ to be that value). $c$ has been measured by looking at how fast light travels, but there are also several other ways of finding $c$. For instance, $c^2$ is equal to the reciprocal of the product of the vacuum permittivity and the vacuum permeability. So not only is the effect of imperfect vacuums negligible in measuring $c$ by looking at the speed of light, but there are multiple other observables that depend on it.
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
add a comment |Â
up vote
12
down vote
An experiment designed to measure some physical quantity such as the speed of light will take into account any perturbing effects. If, for whatever reasons, actually performing speed of light measurements in near vacuum would be impossible, we could still measure it under different air densities and extrapolate the results to zero air density. This extrapolation can be done accurately by fitting the known theoretical dependence of the speed of light on the air density, but we can just as well proceed in a model independent way and not use any theoretical input when doing the extrapolation to a perfect vacuum.
answered Sep 5 at 20:19
Count Iblis
8,08411337
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
add a comment |Â
up vote
0
down vote
The speed of light is by definition exactly 299,792,458 m/s.
If the vacuum was not perfect during our measurements only our definition of a meter would change.
edited Sep 7 at 15:40
safesphere
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
95
down vote
accepted
If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $Delta c/c$ of $3 times 10^-4$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ fracDelta cc = 3 times 10^-4 , P $$
In high vacuum labs we can, without too much effort, get to $10^-10$ torr and this is around $10^-13$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ fracDelta cc approx 3 times 10^-17 $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
answered Sep 5 at 15:55
John Rennie
263k41512761
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
 |Â
show 2 more comments
up vote
95
down vote
accepted
If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $Delta c/c$ of $3 times 10^-4$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ fracDelta cc = 3 times 10^-4 , P $$
In high vacuum labs we can, without too much effort, get to $10^-10$ torr and this is around $10^-13$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ fracDelta cc approx 3 times 10^-17 $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
answered Sep 5 at 15:55
John Rennie
263k41512761
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
 |Â
show 2 more comments
up vote
95
down vote
accepted
up vote
95
down vote
accepted
If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $Delta c/c$ of $3 times 10^-4$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ fracDelta cc = 3 times 10^-4 , P $$
In high vacuum labs we can, without too much effort, get to $10^-10$ torr and this is around $10^-13$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ fracDelta cc approx 3 times 10^-17 $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
answered Sep 5 at 15:55
John Rennie
263k41512761
If we take air, then the refractive index at one atmosphere is around $1.0003$. So if we measure the speed of light in air we get a speed a factor of about $1.0003$ too slow i.e. a fractional error $Delta c/c$ of $3 times 10^-4$.
The difference of the refractive index from one, $n-1$, is proportional to the pressure. Let's write the pressure as a fraction of one atmosphere, i.e. the pressure divided by one atmosphere, then the fractional error in our measurement of $c$ is going to be about:
$$ fracDelta cc = 3 times 10^-4 , P $$
In high vacuum labs we can, without too much effort, get to $10^-10$ torr and this is around $10^-13$ atmospheres or 10 nPa. So measuring the speed of light in this vacuum would give us an error:
$$ fracDelta cc approx 3 times 10^-17 $$
And this is already smaller than the experimental errors in the measurement.
So while it is technically correct that we've never measured the speed of light in a perfect vacuum, the vacuum we can generate is sufficiently good that its effect on the measurement is entirely negligible.
answered Sep 5 at 15:55
John Rennie
263k41512761
edited Sep 6 at 11:49
answered Sep 5 at 15:55
John Rennie
263k41512761
answered Sep 5 at 15:55
John Rennie
263k41512761
answered Sep 5 at 15:55
John Rennie
263k41512761
263k41512761
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
 |Â
show 2 more comments
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
6
6
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
Since $Delta c/c$ is surely dimensionless maybe change the $= 3times 10^-4 P$ to $sim 3times 10^-4 P$?
– ZeroTheHero
Sep 5 at 17:13
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
And if we know the effect of the medium on our measurement then we can correct for that anyway, yes?
– John Bollinger
Sep 5 at 18:35
12
12
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
@ZeroTheHero John explicitly calls for the pressure to be measured in atmospheres, so he gets off on a technicality. But frankly, that should really be expressed as $$fracDelta cc = 3 times 10^-4 fracPP_mathrmatm.$$
– Emilio Pisanty
Sep 5 at 19:36
4
4
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
The phrase "a speed a factor of $3 times 10^-4$ too slow" seems to imply that the unmeasured perfect vacuum speed was about $3333.overline3$ times the speed in air, the actual factor is of course the above $1 + 3 times 10^-4$.
– Leif Willerts
Sep 6 at 11:31
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
@LeifWillerts yes, true, I'm being a little careless about the wording. I'll have a look at tidying that up.
– John Rennie
Sep 6 at 11:39
 |Â
show 2 more comments
up vote
70
down vote
The answer by John Rennie is good so far as the impacts of the imperfect vacuum go, so I won't repeat that here.
As regards the last part of your question about whether this should be accounted when measuring distance, it's worth noting that the standard defines the speed of light to be a specific value and then, using also the definition of the second, derives the meter as a matter of measurement. So as the standards are currently written, the speed of light is exact by definition.
Your question, as written, implicitly assumes that the meter and the second are given by definition and the speed of light a question of measurement.
So from that perspective, your question really should be written to ask whether the impact of imperfect vacuum impacts our definition of the meter. The answer to that, is that it probably does, as was approximately quantified by John Rennie. Whether or not it is important depends on what method is used and what other experimental uncertainties are inherent to that method.
answered Sep 5 at 16:48
Brick
1,194414
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
add a comment |Â
up vote
70
down vote
The answer by John Rennie is good so far as the impacts of the imperfect vacuum go, so I won't repeat that here.
As regards the last part of your question about whether this should be accounted when measuring distance, it's worth noting that the standard defines the speed of light to be a specific value and then, using also the definition of the second, derives the meter as a matter of measurement. So as the standards are currently written, the speed of light is exact by definition.
Your question, as written, implicitly assumes that the meter and the second are given by definition and the speed of light a question of measurement.
So from that perspective, your question really should be written to ask whether the impact of imperfect vacuum impacts our definition of the meter. The answer to that, is that it probably does, as was approximately quantified by John Rennie. Whether or not it is important depends on what method is used and what other experimental uncertainties are inherent to that method.
answered Sep 5 at 16:48
Brick
1,194414
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
add a comment |Â
up vote
70
down vote
up vote
70
down vote
The answer by John Rennie is good so far as the impacts of the imperfect vacuum go, so I won't repeat that here.
As regards the last part of your question about whether this should be accounted when measuring distance, it's worth noting that the standard defines the speed of light to be a specific value and then, using also the definition of the second, derives the meter as a matter of measurement. So as the standards are currently written, the speed of light is exact by definition.
Your question, as written, implicitly assumes that the meter and the second are given by definition and the speed of light a question of measurement.
So from that perspective, your question really should be written to ask whether the impact of imperfect vacuum impacts our definition of the meter. The answer to that, is that it probably does, as was approximately quantified by John Rennie. Whether or not it is important depends on what method is used and what other experimental uncertainties are inherent to that method.
answered Sep 5 at 16:48
Brick
1,194414
The answer by John Rennie is good so far as the impacts of the imperfect vacuum go, so I won't repeat that here.
As regards the last part of your question about whether this should be accounted when measuring distance, it's worth noting that the standard defines the speed of light to be a specific value and then, using also the definition of the second, derives the meter as a matter of measurement. So as the standards are currently written, the speed of light is exact by definition.
Your question, as written, implicitly assumes that the meter and the second are given by definition and the speed of light a question of measurement.
So from that perspective, your question really should be written to ask whether the impact of imperfect vacuum impacts our definition of the meter. The answer to that, is that it probably does, as was approximately quantified by John Rennie. Whether or not it is important depends on what method is used and what other experimental uncertainties are inherent to that method.
answered Sep 5 at 16:48
Brick
1,194414
answered Sep 5 at 16:48
Brick
1,194414
answered Sep 5 at 16:48
Brick
1,194414
answered Sep 5 at 16:48
Brick
1,194414
1,194414
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
add a comment |Â
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
In the old days, the length of a meter was defined from a certain rod of iridium-platinum alloy which is now kept under glass.
– can-ned_food
Sep 7 at 4:23
1
1
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@can-ned_food and the kilogram still is, slowly changing weight.
– Tim
Sep 7 at 20:58
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
@Tim The mass of the standard kilogram can’t change, also by definition.
– Mike Scott
2 days ago
4
4
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
@MikeScott Strictly speaking, it is the value of the standard's mass in units of kg which can't change (since it is fixed to 1 by definition, which I assume is what you meant to say). The mass itself can (and does) change. Just wanted to point out that there is an important difference between a physical quantity and its numerical value in some specific unit.
– aekmr
2 days ago
add a comment |Â
up vote
19
down vote
There is a constant in physics called $c$ that is the "exchange rate" between space and time. One second in time is in some sense "equivalent" to $c$ times one second (which then gives a distance in space). Light is taken to travel at $c$. Note that $c$ isn't the speed of light, but rather the speed of light is $c$, which is a subtle distinction ($c$ being what it is causes light to travel at that speed, rather than light traveling at that speed causes $c$ to be that value). $c$ has been measured by looking at how fast light travels, but there are also several other ways of finding $c$. For instance, $c^2$ is equal to the reciprocal of the product of the vacuum permittivity and the vacuum permeability. So not only is the effect of imperfect vacuums negligible in measuring $c$ by looking at the speed of light, but there are multiple other observables that depend on it.
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
add a comment |Â
up vote
19
down vote
There is a constant in physics called $c$ that is the "exchange rate" between space and time. One second in time is in some sense "equivalent" to $c$ times one second (which then gives a distance in space). Light is taken to travel at $c$. Note that $c$ isn't the speed of light, but rather the speed of light is $c$, which is a subtle distinction ($c$ being what it is causes light to travel at that speed, rather than light traveling at that speed causes $c$ to be that value). $c$ has been measured by looking at how fast light travels, but there are also several other ways of finding $c$. For instance, $c^2$ is equal to the reciprocal of the product of the vacuum permittivity and the vacuum permeability. So not only is the effect of imperfect vacuums negligible in measuring $c$ by looking at the speed of light, but there are multiple other observables that depend on it.
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
add a comment |Â
up vote
19
down vote
up vote
19
down vote
There is a constant in physics called $c$ that is the "exchange rate" between space and time. One second in time is in some sense "equivalent" to $c$ times one second (which then gives a distance in space). Light is taken to travel at $c$. Note that $c$ isn't the speed of light, but rather the speed of light is $c$, which is a subtle distinction ($c$ being what it is causes light to travel at that speed, rather than light traveling at that speed causes $c$ to be that value). $c$ has been measured by looking at how fast light travels, but there are also several other ways of finding $c$. For instance, $c^2$ is equal to the reciprocal of the product of the vacuum permittivity and the vacuum permeability. So not only is the effect of imperfect vacuums negligible in measuring $c$ by looking at the speed of light, but there are multiple other observables that depend on it.
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
There is a constant in physics called $c$ that is the "exchange rate" between space and time. One second in time is in some sense "equivalent" to $c$ times one second (which then gives a distance in space). Light is taken to travel at $c$. Note that $c$ isn't the speed of light, but rather the speed of light is $c$, which is a subtle distinction ($c$ being what it is causes light to travel at that speed, rather than light traveling at that speed causes $c$ to be that value). $c$ has been measured by looking at how fast light travels, but there are also several other ways of finding $c$. For instance, $c^2$ is equal to the reciprocal of the product of the vacuum permittivity and the vacuum permeability. So not only is the effect of imperfect vacuums negligible in measuring $c$ by looking at the speed of light, but there are multiple other observables that depend on it.
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
edited Sep 5 at 19:31
Stéphane Rollandin
2,30031227
2,30031227
answered Sep 5 at 18:52
Acccumulation
1,35719
answered Sep 5 at 18:52
Acccumulation
1,35719
answered Sep 5 at 18:52
Acccumulation
1,35719
1,35719
add a comment |Â
add a comment |Â
up vote
12
down vote
An experiment designed to measure some physical quantity such as the speed of light will take into account any perturbing effects. If, for whatever reasons, actually performing speed of light measurements in near vacuum would be impossible, we could still measure it under different air densities and extrapolate the results to zero air density. This extrapolation can be done accurately by fitting the known theoretical dependence of the speed of light on the air density, but we can just as well proceed in a model independent way and not use any theoretical input when doing the extrapolation to a perfect vacuum.
answered Sep 5 at 20:19
Count Iblis
8,08411337
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
add a comment |Â
up vote
12
down vote
An experiment designed to measure some physical quantity such as the speed of light will take into account any perturbing effects. If, for whatever reasons, actually performing speed of light measurements in near vacuum would be impossible, we could still measure it under different air densities and extrapolate the results to zero air density. This extrapolation can be done accurately by fitting the known theoretical dependence of the speed of light on the air density, but we can just as well proceed in a model independent way and not use any theoretical input when doing the extrapolation to a perfect vacuum.
answered Sep 5 at 20:19
Count Iblis
8,08411337
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
add a comment |Â
up vote
12
down vote
up vote
12
down vote
An experiment designed to measure some physical quantity such as the speed of light will take into account any perturbing effects. If, for whatever reasons, actually performing speed of light measurements in near vacuum would be impossible, we could still measure it under different air densities and extrapolate the results to zero air density. This extrapolation can be done accurately by fitting the known theoretical dependence of the speed of light on the air density, but we can just as well proceed in a model independent way and not use any theoretical input when doing the extrapolation to a perfect vacuum.
answered Sep 5 at 20:19
Count Iblis
8,08411337
An experiment designed to measure some physical quantity such as the speed of light will take into account any perturbing effects. If, for whatever reasons, actually performing speed of light measurements in near vacuum would be impossible, we could still measure it under different air densities and extrapolate the results to zero air density. This extrapolation can be done accurately by fitting the known theoretical dependence of the speed of light on the air density, but we can just as well proceed in a model independent way and not use any theoretical input when doing the extrapolation to a perfect vacuum.
answered Sep 5 at 20:19
Count Iblis
8,08411337
answered Sep 5 at 20:19
Count Iblis
8,08411337
answered Sep 5 at 20:19
Count Iblis
8,08411337
answered Sep 5 at 20:19
Count Iblis
8,08411337
8,08411337
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
add a comment |Â
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
1
1
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
+1 This should be the accepted answer. The actually accepted answer is correct as well, boiling down to "it's an unmeasurable difference", but the fact is that the scientists who (can) do these kinds of measurements would certainly think of any residual matter in outer space and factor that in into their calculations... and this is the aspect the question is shooting for, as far as I can tell...
– AnoE
Sep 7 at 9:05
add a comment |Â
up vote
0
down vote
The speed of light is by definition exactly 299,792,458 m/s.
If the vacuum was not perfect during our measurements only our definition of a meter would change.
edited Sep 7 at 15:40
safesphere
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
add a comment |Â
up vote
0
down vote
The speed of light is by definition exactly 299,792,458 m/s.
If the vacuum was not perfect during our measurements only our definition of a meter would change.
edited Sep 7 at 15:40
safesphere
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The speed of light is by definition exactly 299,792,458 m/s.
If the vacuum was not perfect during our measurements only our definition of a meter would change.
edited Sep 7 at 15:40
safesphere
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
The speed of light is by definition exactly 299,792,458 m/s.
If the vacuum was not perfect during our measurements only our definition of a meter would change.
edited Sep 7 at 15:40
safesphere
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
edited Sep 7 at 15:40
safesphere
6,39111238
edited Sep 7 at 15:40
safesphere
6,39111238
edited Sep 7 at 15:40
safesphere
6,39111238
6,39111238
answered Sep 7 at 10:31
qacwnfq q
909
answered Sep 7 at 10:31
qacwnfq q
909
answered Sep 7 at 10:31
qacwnfq q
909
909
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
add a comment |Â
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
1
1
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
changing the definition of a metre would still change the speed of light. this doesn't actually answer the question because it's just deflecting the apparent effect...
– HyperNeutrino
Sep 7 at 15:56
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
The speed of light in water is approximately 225,000,000 m/s (experimental result). I think the statement should be qualified.
– Peter Mortensen
yesterday
add a comment |Â
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8
@WillihamTotland Only because you chose to display two decimals.
– Mr Lister
Sep 6 at 6:32
8
I think rounding (the already rounded) $2.998 times 10^8 m/s$ to $3 times 10^8 m/s$ is better than stating it as $3.00 times 10^8 m/s$.
– Mick
Sep 6 at 7:36
36
Why bother with approximations? It only takes a few more characters to write the exact value of 299792458 m/s.
– PM 2Ring
Sep 6 at 8:16
8
@mick technically it's not, the .00 is way more precise. $3x10^8$ could even be $3.4$. Saying that it's $3.00x10^8$ is not stating, it's a correct rounding with precise information conveyed. That's what the original comment was about.
– luk32
Sep 6 at 10:06
11
The speed of light in vacuum is exactly $c = 1$.
– Danijel
Sep 6 at 14:03