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How to get Planck length in meters to 6 decimal places
Way to improve “show me this decimal number to M places, don't use scientific notation”?How to display a different number of significant digits in each column of TableForm output?How to get approximate (decimal) output for ArcCosh[e^x]Physical quantities to decimal powerIs it reasonable to have both TemperatureUnit and TemperatureDifferenceUnit?Plotting with units: emits error and gives incorrect plotRemove leading zero and trailing decimal from display of numberRemove decimal point from N functionMax[] is being inconsistent when applied to time seriesHow can I add units to variables inside Manipulate?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I want Mathematica to format quantities to more decimal places than the default.
For example I have tried
NumberForm[
N[
UnitConvert[Quantity["PlanckLength"] , "Meters" ]
,6]
,6, 6]
This is unfortunately incorrect.
I also tried:
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters" ], LongForm"]
but this produces an ugly representation:
I want it to produce an output like the first result (except correct to 6 decimal places).
From Wikipedia and other resources on the web it seems that this quantity is known to 6 decimal places (by default Mathematica truncates it to 4 decimal places.)
output-formatting units
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
I want Mathematica to format quantities to more decimal places than the default.
For example I have tried
NumberForm[
N[
UnitConvert[Quantity["PlanckLength"] , "Meters" ]
,6]
,6, 6]
This is unfortunately incorrect.
I also tried:
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters" ], LongForm"]
but this produces an ugly representation:
I want it to produce an output like the first result (except correct to 6 decimal places).
From Wikipedia and other resources on the web it seems that this quantity is known to 6 decimal places (by default Mathematica truncates it to 4 decimal places.)
output-formatting units
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
$endgroup$
4
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago
add a comment |
$begingroup$
I want Mathematica to format quantities to more decimal places than the default.
For example I have tried
NumberForm[
N[
UnitConvert[Quantity["PlanckLength"] , "Meters" ]
,6]
,6, 6]
This is unfortunately incorrect.
I also tried:
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters" ], LongForm"]
but this produces an ugly representation:
I want it to produce an output like the first result (except correct to 6 decimal places).
From Wikipedia and other resources on the web it seems that this quantity is known to 6 decimal places (by default Mathematica truncates it to 4 decimal places.)
output-formatting units
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
$endgroup$
I want Mathematica to format quantities to more decimal places than the default.
For example I have tried
NumberForm[
N[
UnitConvert[Quantity["PlanckLength"] , "Meters" ]
,6]
,6, 6]
This is unfortunately incorrect.
I also tried:
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters" ], LongForm"]
but this produces an ugly representation:
I want it to produce an output like the first result (except correct to 6 decimal places).
From Wikipedia and other resources on the web it seems that this quantity is known to 6 decimal places (by default Mathematica truncates it to 4 decimal places.)
output-formatting units
output-formatting units
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
edited 1 hour ago
m_goldberg
90.8k8 gold badges75 silver badges204 bronze badges
90.8k8 gold badges75 silver badges204 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
asked 8 hours ago
Conor CosnettConor Cosnett
3,49010 silver badges31 bronze badges
3,49010 silver badges31 bronze badges
4
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago
add a comment |
4
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago
4
4
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The 4.652... at the end of 1.61625500000000006684132`4.652207380644164*^-35 should tell you that Mathematica knows this constant only up to 4.65 decimal digits.
You have to enforce first to treat the number as a higher precision number first:
NumberForm[
SetPrecision[
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"],
"LongForm"], 7],
7, 6]
1.616255 * 10^-35 meters
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
$endgroup$
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.10^-4.652207380644164times the number) in which the actuall value is believed to lie in.
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
add a comment |
$begingroup$
You can use the relative uncertainty 1.1*10^-5 from the CODATA website you also referenced, and use Around to construct a value for the Planck length with uncertainty which has a much nicer formatting for showing the relevant digits of the number:
Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]
(1.616255±0.000018) x 10^-35 m
Since the relative uncertainty is also encoded in the precision of the number in Mathematica like Henrik pointed out, we can also use that to compute the relative uncertainty (the precision of the number representation in Mathematica not always coincides with the measurement uncertainty, but in this specific case it does):
Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.61626±0.00004) x 10^-35 m
which is for some reason off by a factor of two. Correcting that factor gives the same result as earlier:
Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.616255±0.000018) x 10^-35 m
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Ah, very neat to useAround. Have to rembember that (+1).
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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oldest
votes
$begingroup$
The 4.652... at the end of 1.61625500000000006684132`4.652207380644164*^-35 should tell you that Mathematica knows this constant only up to 4.65 decimal digits.
You have to enforce first to treat the number as a higher precision number first:
NumberForm[
SetPrecision[
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"],
"LongForm"], 7],
7, 6]
1.616255 * 10^-35 meters
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
$endgroup$
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.10^-4.652207380644164times the number) in which the actuall value is believed to lie in.
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
add a comment |
$begingroup$
The 4.652... at the end of 1.61625500000000006684132`4.652207380644164*^-35 should tell you that Mathematica knows this constant only up to 4.65 decimal digits.
You have to enforce first to treat the number as a higher precision number first:
NumberForm[
SetPrecision[
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"],
"LongForm"], 7],
7, 6]
1.616255 * 10^-35 meters
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
$endgroup$
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.10^-4.652207380644164times the number) in which the actuall value is believed to lie in.
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
add a comment |
$begingroup$
The 4.652... at the end of 1.61625500000000006684132`4.652207380644164*^-35 should tell you that Mathematica knows this constant only up to 4.65 decimal digits.
You have to enforce first to treat the number as a higher precision number first:
NumberForm[
SetPrecision[
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"],
"LongForm"], 7],
7, 6]
1.616255 * 10^-35 meters
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
$endgroup$
The 4.652... at the end of 1.61625500000000006684132`4.652207380644164*^-35 should tell you that Mathematica knows this constant only up to 4.65 decimal digits.
You have to enforce first to treat the number as a higher precision number first:
NumberForm[
SetPrecision[
QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"],
"LongForm"], 7],
7, 6]
1.616255 * 10^-35 meters
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
66.2k5 gold badges94 silver badges183 bronze badges
66.2k5 gold badges94 silver badges183 bronze badges
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.10^-4.652207380644164times the number) in which the actuall value is believed to lie in.
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
add a comment |
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.10^-4.652207380644164times the number) in which the actuall value is believed to lie in.
$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
$begingroup$
what does "up to 4.65 digits" mean?
$endgroup$
– Conor Cosnett
7 hours ago
1
1
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.
10^-4.652207380644164 times the number) in which the actuall value is believed to lie in.$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
Well, as you might know, floating point numbers are actually stored as binary numbers. "up to 4.65 digits" means that the mantissa stores a certain number of binary digits that leads converts to something close to 4.65 digits. If I am not mistaken, this corresponds to something like 15 to 16 binary digits. But actually, Mathematica's finite precision numbers consist of a number (e.g. 1.61625500000000006684132) and a diameter of an interval (e.g.
10^-4.652207380644164 times the number) in which the actuall value is believed to lie in.$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
$begingroup$
So the "4.65 digits" describe something like a "confidence interval"
$endgroup$
– Henrik Schumacher
6 hours ago
add a comment |
$begingroup$
You can use the relative uncertainty 1.1*10^-5 from the CODATA website you also referenced, and use Around to construct a value for the Planck length with uncertainty which has a much nicer formatting for showing the relevant digits of the number:
Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]
(1.616255±0.000018) x 10^-35 m
Since the relative uncertainty is also encoded in the precision of the number in Mathematica like Henrik pointed out, we can also use that to compute the relative uncertainty (the precision of the number representation in Mathematica not always coincides with the measurement uncertainty, but in this specific case it does):
Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.61626±0.00004) x 10^-35 m
which is for some reason off by a factor of two. Correcting that factor gives the same result as earlier:
Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.616255±0.000018) x 10^-35 m
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Ah, very neat to useAround. Have to rembember that (+1).
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
$begingroup$
You can use the relative uncertainty 1.1*10^-5 from the CODATA website you also referenced, and use Around to construct a value for the Planck length with uncertainty which has a much nicer formatting for showing the relevant digits of the number:
Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]
(1.616255±0.000018) x 10^-35 m
Since the relative uncertainty is also encoded in the precision of the number in Mathematica like Henrik pointed out, we can also use that to compute the relative uncertainty (the precision of the number representation in Mathematica not always coincides with the measurement uncertainty, but in this specific case it does):
Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.61626±0.00004) x 10^-35 m
which is for some reason off by a factor of two. Correcting that factor gives the same result as earlier:
Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.616255±0.000018) x 10^-35 m
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Ah, very neat to useAround. Have to rembember that (+1).
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
$begingroup$
You can use the relative uncertainty 1.1*10^-5 from the CODATA website you also referenced, and use Around to construct a value for the Planck length with uncertainty which has a much nicer formatting for showing the relevant digits of the number:
Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]
(1.616255±0.000018) x 10^-35 m
Since the relative uncertainty is also encoded in the precision of the number in Mathematica like Henrik pointed out, we can also use that to compute the relative uncertainty (the precision of the number representation in Mathematica not always coincides with the measurement uncertainty, but in this specific case it does):
Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.61626±0.00004) x 10^-35 m
which is for some reason off by a factor of two. Correcting that factor gives the same result as earlier:
Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.616255±0.000018) x 10^-35 m
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
$endgroup$
You can use the relative uncertainty 1.1*10^-5 from the CODATA website you also referenced, and use Around to construct a value for the Planck length with uncertainty which has a much nicer formatting for showing the relevant digits of the number:
Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]
(1.616255±0.000018) x 10^-35 m
Since the relative uncertainty is also encoded in the precision of the number in Mathematica like Henrik pointed out, we can also use that to compute the relative uncertainty (the precision of the number representation in Mathematica not always coincides with the measurement uncertainty, but in this specific case it does):
Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.61626±0.00004) x 10^-35 m
which is for some reason off by a factor of two. Correcting that factor gives the same result as earlier:
Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]
(1.616255±0.000018) x 10^-35 m
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
edited 7 hours ago
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
answered 7 hours ago
Thies HeideckeThies Heidecke
7,61126 silver badges39 bronze badges
7,61126 silver badges39 bronze badges
1
$begingroup$
Ah, very neat to useAround. Have to rembember that (+1).
$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
1
$begingroup$
Ah, very neat to useAround. Have to rembember that (+1).
$endgroup$
– Henrik Schumacher
7 hours ago
1
1
$begingroup$
Ah, very neat to use
Around. Have to rembember that (+1).$endgroup$
– Henrik Schumacher
7 hours ago
$begingroup$
Ah, very neat to use
Around. Have to rembember that (+1).$endgroup$
– Henrik Schumacher
7 hours ago
add a comment |
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4
$begingroup$
The Planck length depends on the gravitational constant, which isn't known to ten digits. So even if you forced Mathematica somehow to give you ten digits, they would be meaningless because physically unknown (smaller than the current error bar on the Planck length).
$endgroup$
– Roman
8 hours ago
$begingroup$
I'm voting to close this question as off-topic because it arises from a misconception of physical measurement uncertainty.
$endgroup$
– Roman
8 hours ago
$begingroup$
It seems to be known to 6 decimal places on wikipedia. I will change my question from 10 to 6. (Mathematica only formats it to 4 decimal places)
$endgroup$
– Conor Cosnett
8 hours ago
$begingroup$
This is an instance of a more general question regarding Mathematica's default output form for quantities with units. I chose 10 to accentuate my point. Not because I was interested in knowing the Planck length to 10 decimal places. I hope it is a valid question with 6 decimal places.
$endgroup$
– Conor Cosnett
7 hours ago