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How to get Planck length in meters to 6 decimal places

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2

1

$begingroup$

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

share|improve this question

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 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
  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

1

$begingroup$

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

share|improve this question

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 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
  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 | 

$begingroup$

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

share|improve this question

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 bronze badges

$endgroup$

NumberForm[
 N[
 UnitConvert[Quantity["PlanckLength"] , "Meters" ] 
 ,6] 
 ,6, 6]

QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters" ], LongForm"]

share|improve this question

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 bronze badges

share|improve this question

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 bronze badges

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

edited 1 hour ago

m_goldberg

90.8k88 gold badges7575 silver badges204204 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 bronze badges

asked 8 hours ago

Conor CosnettConor Cosnett

3,4901010 silver badges3131 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

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

share|improve this answer

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 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^-4.652207380644164 times the number) in which the actuall value is believed to lie in.
  $endgroup$
  – 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 | 

6

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, very neat to use Around. Have to rembember that (+1).
  $endgroup$
  – Henrik Schumacher
  6 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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

share|improve this answer

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 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^-4.652207380644164 times the number) in which the actuall value is believed to lie in.
  $endgroup$
  – 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 | 

5

$begingroup$

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

share|improve this answer

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 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^-4.652207380644164 times the number) in which the actuall value is believed to lie in.
  $endgroup$
  – 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 | 

$begingroup$

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

share|improve this answer

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 bronze badges

$endgroup$

NumberForm[
 SetPrecision[
 QuantityForm[UnitConvert[Quantity["PlanckLength"], "Meters"], 
 "LongForm"], 7],
 7, 6]

share|improve this answer

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 bronze badges

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 bronze badges

answered 7 hours ago

Henrik SchumacherHenrik Schumacher

66.2k55 gold badges9494 silver badges183183 bronze badges

6

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, very neat to use Around. Have to rembember that (+1).
  $endgroup$
  – Henrik Schumacher
  6 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, very neat to use Around. Have to rembember that (+1).
  $endgroup$
  – 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

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

$endgroup$

Around[UnitConvert[Quantity["PlanckLength"], "m"], Scaled[1.1*^-5]]

Around[#, Scaled[10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]

Around[#, Scaled[0.5*10^-Precision[#]]] &[UnitConvert[Quantity["PlanckLength"], "m"]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges

answered 7 hours ago

Thies HeideckeThies Heidecke

7,6112626 silver badges3939 bronze badges