How does natural unit make sense?












1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: physics.stackexchange.com/questions/102527/…
    $endgroup$
    – Hanting Zhang
    3 hours ago
















1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: physics.stackexchange.com/questions/102527/…
    $endgroup$
    – Hanting Zhang
    3 hours ago














1












1








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















2












$begingroup$

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 $c$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$


    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






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "151"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f461097%2fhow-does-natural-unit-make-sense%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        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 $c$.






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          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 $c$.






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            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 $c$.






            share|cite|improve this answer











            $endgroup$



            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 $c$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            G. SmithG. Smith

            7,40111425




            7,40111425























                1












                $begingroup$


                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






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$


                  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






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$


                    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






                    share|cite|improve this answer









                    $endgroup$




                    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







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    DaleDale

                    5,7511826




                    5,7511826























                        0












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          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.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            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.






                            share|cite|improve this answer









                            $endgroup$



                            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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            J. MurrayJ. Murray

                            7,7452723




                            7,7452723






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Physics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f461097%2fhow-does-natural-unit-make-sense%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Loup dans la culture

                                How to solve the problem of ntp “Unable to contact time server” from KDE?

                                ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux