How does natural unit make sense?
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Both the fundamental constants $hbar$ and $c$ have dimensions. In particular, $[hbar]=ML^2T^{-1}$ and $[c]=LT^{-1}$. But in natural units, we make them dimensionless constants of equal magnitude. How is this possible? This means length is measured in the same units as time which is measured in the units of inverse mass! Am I measuring the distance between two points in seconds??? Can we do that? I cannot make myself comfortable with it :-( Also why is it not possible to make $hbar=c=G=1$ simultaneously?
particle-physics units dimensional-analysis si-units
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add a comment |
$begingroup$
Both the fundamental constants $hbar$ and $c$ have dimensions. In particular, $[hbar]=ML^2T^{-1}$ and $[c]=LT^{-1}$. But in natural units, we make them dimensionless constants of equal magnitude. How is this possible? This means length is measured in the same units as time which is measured in the units of inverse mass! Am I measuring the distance between two points in seconds??? Can we do that? I cannot make myself comfortable with it :-( Also why is it not possible to make $hbar=c=G=1$ simultaneously?
particle-physics units dimensional-analysis si-units
$endgroup$
$begingroup$
Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago
add a comment |
$begingroup$
Both the fundamental constants $hbar$ and $c$ have dimensions. In particular, $[hbar]=ML^2T^{-1}$ and $[c]=LT^{-1}$. But in natural units, we make them dimensionless constants of equal magnitude. How is this possible? This means length is measured in the same units as time which is measured in the units of inverse mass! Am I measuring the distance between two points in seconds??? Can we do that? I cannot make myself comfortable with it :-( Also why is it not possible to make $hbar=c=G=1$ simultaneously?
particle-physics units dimensional-analysis si-units
$endgroup$
Both the fundamental constants $hbar$ and $c$ have dimensions. In particular, $[hbar]=ML^2T^{-1}$ and $[c]=LT^{-1}$. But in natural units, we make them dimensionless constants of equal magnitude. How is this possible? This means length is measured in the same units as time which is measured in the units of inverse mass! Am I measuring the distance between two points in seconds??? Can we do that? I cannot make myself comfortable with it :-( Also why is it not possible to make $hbar=c=G=1$ simultaneously?
particle-physics units dimensional-analysis si-units
particle-physics units dimensional-analysis si-units
edited 3 hours ago
mithusengupta123
asked 3 hours ago
mithusengupta123mithusengupta123
1,23711435
1,23711435
$begingroup$
Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago
add a comment |
$begingroup$
Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago
$begingroup$
Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago
$begingroup$
Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Can we measure distance in seconds? Definitely. When you set $c=1$, “one second of distance” is simply the distance that light travels in vacuum in one second. And “one meter of time” is the amount of time it takes for light to go one meter in vacuum.
It is possible to set $hbar=c=G=1$ simultaneously. This is what produces Planck units.
Natural units are not just a nice simplification. They allow the physics of an equation to “shine through” more clearly. For example,
$$E^2-mathbf{p}^2=m^2$$
says “Mass is the invariant length of the energy-momentum four vector” more clearly than
$$E^2-mathbf{p}^2 c^2=m^2 c^4$$
does, with its extraneous factors of $c$.
$endgroup$
add a comment |
$begingroup$
But in natural units, we make them dimensionless constants of equal magnitude. How is this possible?
It is not widely known or appreciated, but both the magnitude and the dimensionality of the units we use are a matter of convention. We are free to use different unit systems with different conventions, as long as we use them self consistently. The most important thing to remember is that the laws of physics are slightly different if we write them in other units.
CGS units are the most widely used units with substantial differences in dimensionality compared to SI. For example, in SI units the Coulomb is a base unit with dimensions of $Q$, but in CGS units the statcoulomb is a derived unit with dimensions of $L^{3/2}M^{1/2}T^{-1}$. As a result, Coulomb’s law takes a particularly simple form: $$F=frac{q_1 q_2}{r^2}$$
Am I measuring the distance between two points in seconds??? Can we do that?
Yes, but not in SI units. There are also units where time would have dimensions of length. Those are also valid. You simply have to adapt the equations appropriately.
Also why is it not possible to make ℏ=𝑐=𝐺=1 simultaneously?
It is. One example is Planck units https://en.m.wikipedia.org/wiki/Planck_units
$endgroup$
add a comment |
$begingroup$
The other answers more directly address your questions, but I would point out that many people casually measure distance and time in the same units. If you've ever said "I live twenty minutes away from my office," then you've used units in which the characteristic speed (i.e. the conversion factor between distance and time - in this case, the average speed of traffic) has been set to 1.
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Can we measure distance in seconds? Definitely. When you set $c=1$, “one second of distance” is simply the distance that light travels in vacuum in one second. And “one meter of time” is the amount of time it takes for light to go one meter in vacuum.
It is possible to set $hbar=c=G=1$ simultaneously. This is what produces Planck units.
Natural units are not just a nice simplification. They allow the physics of an equation to “shine through” more clearly. For example,
$$E^2-mathbf{p}^2=m^2$$
says “Mass is the invariant length of the energy-momentum four vector” more clearly than
$$E^2-mathbf{p}^2 c^2=m^2 c^4$$
does, with its extraneous factors of $c$.
$endgroup$
add a comment |
$begingroup$
Can we measure distance in seconds? Definitely. When you set $c=1$, “one second of distance” is simply the distance that light travels in vacuum in one second. And “one meter of time” is the amount of time it takes for light to go one meter in vacuum.
It is possible to set $hbar=c=G=1$ simultaneously. This is what produces Planck units.
Natural units are not just a nice simplification. They allow the physics of an equation to “shine through” more clearly. For example,
$$E^2-mathbf{p}^2=m^2$$
says “Mass is the invariant length of the energy-momentum four vector” more clearly than
$$E^2-mathbf{p}^2 c^2=m^2 c^4$$
does, with its extraneous factors of $c$.
$endgroup$
add a comment |
$begingroup$
Can we measure distance in seconds? Definitely. When you set $c=1$, “one second of distance” is simply the distance that light travels in vacuum in one second. And “one meter of time” is the amount of time it takes for light to go one meter in vacuum.
It is possible to set $hbar=c=G=1$ simultaneously. This is what produces Planck units.
Natural units are not just a nice simplification. They allow the physics of an equation to “shine through” more clearly. For example,
$$E^2-mathbf{p}^2=m^2$$
says “Mass is the invariant length of the energy-momentum four vector” more clearly than
$$E^2-mathbf{p}^2 c^2=m^2 c^4$$
does, with its extraneous factors of $c$.
$endgroup$
Can we measure distance in seconds? Definitely. When you set $c=1$, “one second of distance” is simply the distance that light travels in vacuum in one second. And “one meter of time” is the amount of time it takes for light to go one meter in vacuum.
It is possible to set $hbar=c=G=1$ simultaneously. This is what produces Planck units.
Natural units are not just a nice simplification. They allow the physics of an equation to “shine through” more clearly. For example,
$$E^2-mathbf{p}^2=m^2$$
says “Mass is the invariant length of the energy-momentum four vector” more clearly than
$$E^2-mathbf{p}^2 c^2=m^2 c^4$$
does, with its extraneous factors of $c$.
edited 3 hours ago
answered 3 hours ago
G. SmithG. Smith
7,40111425
7,40111425
add a comment |
add a comment |
$begingroup$
But in natural units, we make them dimensionless constants of equal magnitude. How is this possible?
It is not widely known or appreciated, but both the magnitude and the dimensionality of the units we use are a matter of convention. We are free to use different unit systems with different conventions, as long as we use them self consistently. The most important thing to remember is that the laws of physics are slightly different if we write them in other units.
CGS units are the most widely used units with substantial differences in dimensionality compared to SI. For example, in SI units the Coulomb is a base unit with dimensions of $Q$, but in CGS units the statcoulomb is a derived unit with dimensions of $L^{3/2}M^{1/2}T^{-1}$. As a result, Coulomb’s law takes a particularly simple form: $$F=frac{q_1 q_2}{r^2}$$
Am I measuring the distance between two points in seconds??? Can we do that?
Yes, but not in SI units. There are also units where time would have dimensions of length. Those are also valid. You simply have to adapt the equations appropriately.
Also why is it not possible to make ℏ=𝑐=𝐺=1 simultaneously?
It is. One example is Planck units https://en.m.wikipedia.org/wiki/Planck_units
$endgroup$
add a comment |
$begingroup$
But in natural units, we make them dimensionless constants of equal magnitude. How is this possible?
It is not widely known or appreciated, but both the magnitude and the dimensionality of the units we use are a matter of convention. We are free to use different unit systems with different conventions, as long as we use them self consistently. The most important thing to remember is that the laws of physics are slightly different if we write them in other units.
CGS units are the most widely used units with substantial differences in dimensionality compared to SI. For example, in SI units the Coulomb is a base unit with dimensions of $Q$, but in CGS units the statcoulomb is a derived unit with dimensions of $L^{3/2}M^{1/2}T^{-1}$. As a result, Coulomb’s law takes a particularly simple form: $$F=frac{q_1 q_2}{r^2}$$
Am I measuring the distance between two points in seconds??? Can we do that?
Yes, but not in SI units. There are also units where time would have dimensions of length. Those are also valid. You simply have to adapt the equations appropriately.
Also why is it not possible to make ℏ=𝑐=𝐺=1 simultaneously?
It is. One example is Planck units https://en.m.wikipedia.org/wiki/Planck_units
$endgroup$
add a comment |
$begingroup$
But in natural units, we make them dimensionless constants of equal magnitude. How is this possible?
It is not widely known or appreciated, but both the magnitude and the dimensionality of the units we use are a matter of convention. We are free to use different unit systems with different conventions, as long as we use them self consistently. The most important thing to remember is that the laws of physics are slightly different if we write them in other units.
CGS units are the most widely used units with substantial differences in dimensionality compared to SI. For example, in SI units the Coulomb is a base unit with dimensions of $Q$, but in CGS units the statcoulomb is a derived unit with dimensions of $L^{3/2}M^{1/2}T^{-1}$. As a result, Coulomb’s law takes a particularly simple form: $$F=frac{q_1 q_2}{r^2}$$
Am I measuring the distance between two points in seconds??? Can we do that?
Yes, but not in SI units. There are also units where time would have dimensions of length. Those are also valid. You simply have to adapt the equations appropriately.
Also why is it not possible to make ℏ=𝑐=𝐺=1 simultaneously?
It is. One example is Planck units https://en.m.wikipedia.org/wiki/Planck_units
$endgroup$
But in natural units, we make them dimensionless constants of equal magnitude. How is this possible?
It is not widely known or appreciated, but both the magnitude and the dimensionality of the units we use are a matter of convention. We are free to use different unit systems with different conventions, as long as we use them self consistently. The most important thing to remember is that the laws of physics are slightly different if we write them in other units.
CGS units are the most widely used units with substantial differences in dimensionality compared to SI. For example, in SI units the Coulomb is a base unit with dimensions of $Q$, but in CGS units the statcoulomb is a derived unit with dimensions of $L^{3/2}M^{1/2}T^{-1}$. As a result, Coulomb’s law takes a particularly simple form: $$F=frac{q_1 q_2}{r^2}$$
Am I measuring the distance between two points in seconds??? Can we do that?
Yes, but not in SI units. There are also units where time would have dimensions of length. Those are also valid. You simply have to adapt the equations appropriately.
Also why is it not possible to make ℏ=𝑐=𝐺=1 simultaneously?
It is. One example is Planck units https://en.m.wikipedia.org/wiki/Planck_units
answered 3 hours ago
DaleDale
5,7511826
5,7511826
add a comment |
add a comment |
$begingroup$
The other answers more directly address your questions, but I would point out that many people casually measure distance and time in the same units. If you've ever said "I live twenty minutes away from my office," then you've used units in which the characteristic speed (i.e. the conversion factor between distance and time - in this case, the average speed of traffic) has been set to 1.
$endgroup$
add a comment |
$begingroup$
The other answers more directly address your questions, but I would point out that many people casually measure distance and time in the same units. If you've ever said "I live twenty minutes away from my office," then you've used units in which the characteristic speed (i.e. the conversion factor between distance and time - in this case, the average speed of traffic) has been set to 1.
$endgroup$
add a comment |
$begingroup$
The other answers more directly address your questions, but I would point out that many people casually measure distance and time in the same units. If you've ever said "I live twenty minutes away from my office," then you've used units in which the characteristic speed (i.e. the conversion factor between distance and time - in this case, the average speed of traffic) has been set to 1.
$endgroup$
The other answers more directly address your questions, but I would point out that many people casually measure distance and time in the same units. If you've ever said "I live twenty minutes away from my office," then you've used units in which the characteristic speed (i.e. the conversion factor between distance and time - in this case, the average speed of traffic) has been set to 1.
answered 2 hours ago
J. MurrayJ. Murray
7,7452723
7,7452723
add a comment |
add a comment |
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Related: physics.stackexchange.com/questions/102527/…
$endgroup$
– Hanting Zhang
3 hours ago