Universal time, mass, length

5953 /www/nrich/html/content/id/5953/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/6504">Warm-up</a></li> <li><a href="http://nrich.maths.org/6153">Try this next</a></li> <li><a href="http://nrich.maths.org/5838">Think higher</a></li> <li><a href="http://nrich.maths.org/6636">Read: mathematics</a></li> <li><a href="http://plus.maths.org/content/outer-space-are-constants-nature-really-constant">Read: science</a></li> <li><a href="http://plus.maths.org/content/science-fiction-science-fact-reports-frontiers-physics">Explore further</a></li> </ul> <div> </div> The laws of physics involve certain fundamental constants of nature. The key constants which the universe knows about are: <table style="" border="1"> <tbody> <tr> <td>Name</td> <td>Symbol</td> <td>Value</td> </tr> <tr> <td>Newton's gravitational constant</td> <td>$G$</td> <td>$6.674 \times 10^{-8} cm^3\,g^{-1}\,s^{-2}$</td> </tr> <tr> <td>The speed of light</td> <td>$c$</td> <td>$2.998 \times 10^{10} \,cm\,s^{-1}$</td> </tr> <tr> <td>Planck's constant</td> <td>$h$</td> <td>$1.054\times 10^{-27} \,g \,cm^2\,s^{-1}$</td> </tr> <tr> <td>Boltzmann's constant</td> <td>$k$</td> <td>$1.38 \times 10^{-23} \,J\,K^{-1}$</td> </tr> </tbody> </table> What values would these constants have taken in the days of the British Empire, when units of feet (ft) for length, pounds (lb) for weight, calories (cal) for heat and degrees Fahrenheit (F) for temperature, were used? You may use these conversions: <ul> <li>1 cal = 4.184 J</li> <li>2.2 lb = 1kg</li> <li>1 ft = 30.48 cm</li> <li>Change of 1 K = Change of 1.8 degrees F</li> </ul> What units would you need to choose so that the first three constants take the value 1? These are the natural length, mass and time scales for the universe. <div class="framework"> NOTES AND BACKGROUND Whilst the choice of units in most problems is arbitrary, and therefore might seem largely irrelevant, it is important to note that independence of physics on units is actually highly significant and leads to the concept of scale invariance and the powerful tool of dimensional analysis, which is used in the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=5838&part=">Dam Busters 2</a> . Mathematically, the concepts of scale invariance are modelled by fractals on which the details of a system appear qualitatively the same at any order of magnification. Scale invariant systems have interesting statistical properties. This is explored in the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=5937&part=">Scale Invariance</a></div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We have had two very impressive, and beautifully presented, solutions to this problems, from <a class="editorial" href="/content/id/5953/lauraConstants.doc">Laura</a> , from The Henrietta Barnett School and <a class="editorial" href="/content/id/5953/StevenConstants.doc">Steven</a> , from City of Sunderland College. We need the values of 1cm, 1s, 1J and 1K in terms of imperial units. These are $$ 1cm = \frac{1}{30.48}ft\quad1g = \frac{2.2}{1000}lb\quad 1J = \frac{1}{4.184} cal\quad 1K = 1.8^\circ F $$ These values can now be substituted into the numerical expressions for $G, h, c, k$. For Newton's constant we have $$ \begin{eqnarray} G& =& 6.674 \times 10^{-8} cm^3 g^{-1} s^{-2}\\ &=& 6.674 \times 10^{-8} \left(\frac{1}{30.48}ft\right)^3\left(\frac{2.2}{1000}lb\right)^{-1}s^{-2}\\ &=& \frac{6.674\times 10^{-8}\times 1000}{30.48^3\times 2.2}ft^3 lb^{-1} s^{-2}\\ &=& 1.071\times 10^{-9}ft^3lb^{-1}s^{-2} \end{eqnarray} $$ In a similar way, we can find that $$ c = 9.836\times 10^8 ft s^{-1}\quad \quad h = 2.496\times 10^{-33} lb ft^2 s^{-1} \quad\quad k = 1.83\times 10^{-24}cal ^\circ F$$ For the second part of the problem, Steven noted the following In a system of units where $G, c, h, k$ are numerically equal to 1, let $L, M, T$ be the units of length, mass and time respectively, so that $$ \begin{eqnarray} G&=& 1 L^3M^{-1}T^{-2}\\ c&=& 1LT^{-1}\\ h&=& 1ML^2 T^{-1} \end{eqnarray} $$ We now need to solve for $M, L, T$ in terms of $c, G, h$ After some rearranging, Steven found out that $$ M=\sqrt{\frac{ch}{G}}\quad\quad L=\sqrt{\frac{Gh}{c^3}}\quad\quad T=\sqrt{\frac{Gh}{c^5}} $$ Putting in the numerical values given in the question yielded $$ M = 2.176\times 10^{-5}g\quad\quad L = 1.616\times 10^{-33}cm\quad\quad T = 5.389\times 10^{-44}s $$ Steven finally concluded with a clever extension in which he found a natural temperature scale for the universe as follows: From the equation Work = Force x Distance it can be seen that $J = Nm$, and from the equation Force = Mass x Acceleration we have thatn $N = kg m s^{-2}$. So, the units of the Bolzmann constant are $kgm^2s^{-2}K^{-1}$ So, as before, supposing the $\theta$ is the natural temperature scale for the universe $$ k = 1 ML^2T^{-2}\theta^{-1} $$ Solving as before gives $$ \theta = \sqrt{\frac{c^5h}{Gk^2}} $$ The Boltzmann constant must be converted to have the same units as the other constants before evaluating this result. Putting in the numbers gives $$ k = 1.38\times10^{-16}g cm^2 s^{-2}K^{-1} $$ We can now substitute this value into the expression for $\theta$ to give $$ \theta = 1.42\times 10^{32}K $$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> This problem places units in an interesting context which can lead on to other problems. It requires relatively little technical skill, allowing students to focus solely on the process of changing units. <h3>Possible approach</h3> Discuss the problem as a class. Is everyone familiar with each of the units mentioned? Throughout, encourage numerical order of magnitude checks of the answers. <h3>Key questions</h3> <ul> <li>Before changing units, do you expect the answer to be numerically larger or smaller after the change?</li> <li>Relative to the natural scales of the universe are we small, heavy, slow?</li> </ul> <h3>Possible extension</h3> Consider the meaning of the statement 'the natural scales for the universe'. Relate these units to everyday objects. Work out the ratios for a person / a proton/ the sun / the galaxy / a neutron star in the natural units. How do they compare G, c, and h? <h3>Possible support</h3> You could treat this simply as a change of units question. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> When changing units we just need to multiply or divide by a scaling factor. You can use common sense to decide whether to multiply or divide. </mdoxml> 2 3 0 0 0 0 1 Universal time, mass, length Can you work out the natural time scale for the universe? 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