Why are the units of angular acceleration the same as that of angular velocity squared?

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According to this answer, the units for angular velocity squared are $mathrmrad/s^2$. The units for angular acceleration are also $mathrmrad/s^2$. Why is this the case?










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  • 4




    Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
    – Chemomechanics
    Dec 5 at 15:55











  • Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
    – knzhou
    Dec 5 at 16:44










  • Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
    – Cort Ammon
    Dec 5 at 20:12










  • In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
    – Cort Ammon
    Dec 5 at 20:12















up vote
6
down vote

favorite












According to this answer, the units for angular velocity squared are $mathrmrad/s^2$. The units for angular acceleration are also $mathrmrad/s^2$. Why is this the case?










share|cite|improve this question



















  • 4




    Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
    – Chemomechanics
    Dec 5 at 15:55











  • Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
    – knzhou
    Dec 5 at 16:44










  • Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
    – Cort Ammon
    Dec 5 at 20:12










  • In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
    – Cort Ammon
    Dec 5 at 20:12













up vote
6
down vote

favorite









up vote
6
down vote

favorite











According to this answer, the units for angular velocity squared are $mathrmrad/s^2$. The units for angular acceleration are also $mathrmrad/s^2$. Why is this the case?










share|cite|improve this question















According to this answer, the units for angular velocity squared are $mathrmrad/s^2$. The units for angular acceleration are also $mathrmrad/s^2$. Why is this the case?







acceleration units angular-velocity






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edited Dec 5 at 20:57









Ruslan

8,61842868




8,61842868










asked Dec 5 at 15:42









Raymo111

1397




1397







  • 4




    Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
    – Chemomechanics
    Dec 5 at 15:55











  • Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
    – knzhou
    Dec 5 at 16:44










  • Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
    – Cort Ammon
    Dec 5 at 20:12










  • In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
    – Cort Ammon
    Dec 5 at 20:12













  • 4




    Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
    – Chemomechanics
    Dec 5 at 15:55











  • Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
    – knzhou
    Dec 5 at 16:44










  • Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
    – Cort Ammon
    Dec 5 at 20:12










  • In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
    – Cort Ammon
    Dec 5 at 20:12








4




4




Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
– Chemomechanics
Dec 5 at 15:55





Please edit your question to match the details of that answer, which says that dimensionally, both quantities are taken as $1/s^2$. It does not say that the units of angular velocity squared are $mathrmrad/s^2$ (they would be $(mathrmrad/s)^2$).
– Chemomechanics
Dec 5 at 15:55













Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
– knzhou
Dec 5 at 16:44




Units don't tell you much. There are hundreds of interesting physical quantities out there and only $3$ independent fundamental units, so of course sometimes units will match.
– knzhou
Dec 5 at 16:44












Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
– Cort Ammon
Dec 5 at 20:12




Something I found helpful but is not often taught: In math, we do not need a unit for "radians" because it is just defined by a ratio of length. Thus mathematically, radians must be dimensionless. However, in practical physics and engineering situations, retaining a "rad" unit, even though you didn't need it, turns out to be helpful for catching errors. Most of the time radians have to cancel, or be passed to a sin/cos/tan function.
– Cort Ammon
Dec 5 at 20:12












In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
– Cort Ammon
Dec 5 at 20:12





In the few times where it doesn't, such as the small angle approximation $sin(theta) approx theta$, its nice to have a reminder that you're doing something interesting that's worth taking a second look at to make sure it's legitimate.
– Cort Ammon
Dec 5 at 20:12











5 Answers
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up vote
19
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They have the same units of $mathrm s^-2$ only if you don't use $mathrmrad$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrmrad/mathrm s^2$, and that of square of angular velocity is $mathrmrad^2/mathrm s^2$.



If radian is dimensionless, why do we not "simplify" $mathrmrad^2$ to $mathrmrad$? For the same reason as why we introduced $mathrmrad$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrmrad^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.






share|cite|improve this answer





























    up vote
    3
    down vote













    By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.



    Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.






    share|cite|improve this answer



























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      0
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      They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^-2$ and when you derive you get the same.






      share|cite|improve this answer


















      • 4




        Yes, but my question was about why they were the same, not if they were the same.
        – Raymo111
        Dec 5 at 15:49

















      up vote
      0
      down vote













      Radians are dimensionless. You can safely set $textradto 1$. Using $[x]$ to mean "the units of $x$, we find:



      The units of angular velocity $[omega] = left[fracdthetadtright]= frac1T$ where $T$ is time. Then $[omega^2] = frac1T^2$.



      The units of angular acceleration $[alpha] = left[fracd^2thetadt^2right] = frac1T^2$.






      share|cite|improve this answer



























        up vote
        0
        down vote













        Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).



        Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.



        We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.






        share|cite|improve this answer




















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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          19
          down vote



          accepted










          They have the same units of $mathrm s^-2$ only if you don't use $mathrmrad$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
          But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrmrad/mathrm s^2$, and that of square of angular velocity is $mathrmrad^2/mathrm s^2$.



          If radian is dimensionless, why do we not "simplify" $mathrmrad^2$ to $mathrmrad$? For the same reason as why we introduced $mathrmrad$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrmrad^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.






          share|cite|improve this answer


























            up vote
            19
            down vote



            accepted










            They have the same units of $mathrm s^-2$ only if you don't use $mathrmrad$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
            But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrmrad/mathrm s^2$, and that of square of angular velocity is $mathrmrad^2/mathrm s^2$.



            If radian is dimensionless, why do we not "simplify" $mathrmrad^2$ to $mathrmrad$? For the same reason as why we introduced $mathrmrad$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrmrad^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.






            share|cite|improve this answer
























              up vote
              19
              down vote



              accepted







              up vote
              19
              down vote



              accepted






              They have the same units of $mathrm s^-2$ only if you don't use $mathrmrad$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
              But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrmrad/mathrm s^2$, and that of square of angular velocity is $mathrmrad^2/mathrm s^2$.



              If radian is dimensionless, why do we not "simplify" $mathrmrad^2$ to $mathrmrad$? For the same reason as why we introduced $mathrmrad$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrmrad^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.






              share|cite|improve this answer














              They have the same units of $mathrm s^-2$ only if you don't use $mathrmrad$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units).
              But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $mathrmrad/mathrm s^2$, and that of square of angular velocity is $mathrmrad^2/mathrm s^2$.



              If radian is dimensionless, why do we not "simplify" $mathrmrad^2$ to $mathrmrad$? For the same reason as why we introduced $mathrmrad$ in the first place: it's not required to use this symbol, but it does help us remember that we have an angle somewhere. Similarly, if we square it, we must use $mathrmrad^2$ because now we have a square of that angle. But as the unit is dimensionless, it's technically not necessary. It's simply for convenience. We could invent a bunch of other dimensionless units to aid us in bookkeeping, but once we've done it, we must keep their correct powers, otherwise these units are simply useless.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 5 at 20:55

























              answered Dec 5 at 16:10









              Ruslan

              8,61842868




              8,61842868




















                  up vote
                  3
                  down vote













                  By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.



                  Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.






                  share|cite|improve this answer
























                    up vote
                    3
                    down vote













                    By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.



                    Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.






                    share|cite|improve this answer






















                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.



                      Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.






                      share|cite|improve this answer












                      By definition, angular velocity is defined as the rate of change of angle with respect to time, leading to the equation $omega = Delta theta / Delta t$. From dimensional analysis, this yields units of radians/s. Also by definition, angular acceleration is defined as the rate of change of angular velocity with respect to time, leading to the equation $alpha = Delta omega / Delta t$. From dimensional analysis, $Delta omega$ has units of rad/s, so the units of $alpha$ are $rad/s * 1/s$, leading to final units of $rad/s^2$.



                      Note that directly comparing units of $omega ^2$ to units of $alpha$ with no physical basis for doing so, doesn't make sense from a physics standpoint.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 5 at 16:22









                      David White

                      3,8731519




                      3,8731519




















                          up vote
                          0
                          down vote













                          They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^-2$ and when you derive you get the same.






                          share|cite|improve this answer


















                          • 4




                            Yes, but my question was about why they were the same, not if they were the same.
                            – Raymo111
                            Dec 5 at 15:49














                          up vote
                          0
                          down vote













                          They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^-2$ and when you derive you get the same.






                          share|cite|improve this answer


















                          • 4




                            Yes, but my question was about why they were the same, not if they were the same.
                            – Raymo111
                            Dec 5 at 15:49












                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^-2$ and when you derive you get the same.






                          share|cite|improve this answer














                          They have the same dimensions because radians are adimensional. The only physical dimension is time so when you square it you get a $time^-2$ and when you derive you get the same.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 5 at 16:04

























                          answered Dec 5 at 15:48









                          Run like hell

                          986524




                          986524







                          • 4




                            Yes, but my question was about why they were the same, not if they were the same.
                            – Raymo111
                            Dec 5 at 15:49












                          • 4




                            Yes, but my question was about why they were the same, not if they were the same.
                            – Raymo111
                            Dec 5 at 15:49







                          4




                          4




                          Yes, but my question was about why they were the same, not if they were the same.
                          – Raymo111
                          Dec 5 at 15:49




                          Yes, but my question was about why they were the same, not if they were the same.
                          – Raymo111
                          Dec 5 at 15:49










                          up vote
                          0
                          down vote













                          Radians are dimensionless. You can safely set $textradto 1$. Using $[x]$ to mean "the units of $x$, we find:



                          The units of angular velocity $[omega] = left[fracdthetadtright]= frac1T$ where $T$ is time. Then $[omega^2] = frac1T^2$.



                          The units of angular acceleration $[alpha] = left[fracd^2thetadt^2right] = frac1T^2$.






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote













                            Radians are dimensionless. You can safely set $textradto 1$. Using $[x]$ to mean "the units of $x$, we find:



                            The units of angular velocity $[omega] = left[fracdthetadtright]= frac1T$ where $T$ is time. Then $[omega^2] = frac1T^2$.



                            The units of angular acceleration $[alpha] = left[fracd^2thetadt^2right] = frac1T^2$.






                            share|cite|improve this answer






















                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Radians are dimensionless. You can safely set $textradto 1$. Using $[x]$ to mean "the units of $x$, we find:



                              The units of angular velocity $[omega] = left[fracdthetadtright]= frac1T$ where $T$ is time. Then $[omega^2] = frac1T^2$.



                              The units of angular acceleration $[alpha] = left[fracd^2thetadt^2right] = frac1T^2$.






                              share|cite|improve this answer












                              Radians are dimensionless. You can safely set $textradto 1$. Using $[x]$ to mean "the units of $x$, we find:



                              The units of angular velocity $[omega] = left[fracdthetadtright]= frac1T$ where $T$ is time. Then $[omega^2] = frac1T^2$.



                              The units of angular acceleration $[alpha] = left[fracd^2thetadt^2right] = frac1T^2$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 6 at 3:39









                              zahbaz

                              382316




                              382316




















                                  up vote
                                  0
                                  down vote













                                  Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).



                                  Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.



                                  We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.






                                  share|cite|improve this answer
























                                    up vote
                                    0
                                    down vote













                                    Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).



                                    Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.



                                    We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.






                                    share|cite|improve this answer






















                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).



                                      Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.



                                      We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.






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                                      Your question is a bit like why does my decimal representation of 1/3 never terminate? It's because we have chosen base 10 and if we choose a different base (e.g. 9) we can make 1/3 terminate (e.g. 0.3).



                                      Units are just comparing a quantity to a reference quantity chosen by convention. I could use seconds as a distance measurement, where it was understood that the reference distance was how far light travels in a second. In that case distance and time have the same units and distance becomes dimensionless akin to refractive index.



                                      We choose units to keep track of what references we have used, hence the use of rad even when it is dimensionless, and also why we use metres instead of seconds for distance. Units are much more conventional than we often take them to be. Look at the rash of unit systems used in electro-magnetism.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 6 at 12:36









                                      geoff22873

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