Imibuzo Eyisibonelo Exoxa Ngomthelela Wokuhlobana Kuka-Einstein
Imibono ka-Einstein yokuhlobana, okuhlanganisa nemibono yakhe ekhethekile nejwayelekile yokuhlobana, ishintshe ukuqonda kwethu isikhala, isikhathi, kanye namandla adonsela phansi. Nakuba u-Einstein aqala ukwethula le mibono ekuqaleni kwekhulu lama-20, umthelela wayo kwisayensi yesimanje kanye nobuchwepheshe ube mkhulu kakhulu. Lesi sihloko sizohlola izibonelo eziningana zezinkinga ezihlola umthelela obalulekile wokuhlobana kuka-Einstein ezimweni ezahlukene futhi sibonise ukuthi le mbono iguqule kanjani indlela yethu yesayensi.
Isibonelo Umbuzo 1: Ukwanda Kwesikhathi Nokuhamba Emkhathini
Umbuzo:
I-astronaut iya enkanyezini eqhele ngeminyaka yokukhanya engu-4 ukusuka eMhlabeni ngesivinini sokukhanya esiphindwe izikhathi ezingu-0,8 (0,8c). I-astronaut icabanga ukuthi lolu hambo luzothatha isikhathi esingakanani?
Ingxoxo:
Ukuze siqonde isimo sokwanda kwesikhathi, sisebenzisa ifomula eyisisekelo yokuhlobana okukhethekile:
\[t' = \frac{t}{\gamma} \]
lapho \( \gamma \) kuyisici seLorentz esinikezwe ngu:
\[ \gamma = \frac{1}{\sqrt{1 – \left(\frac{v}{c}\right)^2}} \]
Lapha, \( v = 0,8c \) kanye \( c \) yijubane lokukhanya. Bese kuthi,
\[ \gamma = \frac{1}{\sqrt{1 – (0,8)^2}} = \frac{1}{\sqrt{1 – 0,64}} = \frac{1}{\sqrt{0,36}} = \frac{1}{0,6} \cishe kube ngu-1,667 \]
Uma ibanga lenkanyezi liyiminyaka yokukhanya engu-4 futhi i-astronaut ihamba ngesivinini esingu-0,8c, isikhathi esibonwe ngumuntu obuka eMhlabeni (t) yilesi:
\[ t = \frac{Ibanga}{Isivinini} = \frac{4 \umbhalo{iminyaka yokukhanya}}{0,8c} = 5 \umbhalo{iminyaka} \]
Kodwa-ke, isikhathi esibonwa yi-astronaut (t') yilesi:
\[ t' = \frac{t}{\gamma} = \frac{5 \text{ years}}{1,667} \cishe kube ngu-3 \text{ years} \]
Ngakho-ke, ngokusho kosomkhathi, lolu hambo luthathe iminyaka engaba mi-3 kuphela, yize ngokombono woMhlaba luthathe iminyaka emi-5.
Isibonelo Umbuzo 2: Ukufinyela Ubude kanye Nokubuka Kokuhlola
Umbuzo:
Indizamkhathi ingamamitha ayi-100 ubude uma iphumula uma iqhathaniswa noMhlaba. Uma indizamkhathi ihamba ngesivinini esingu-0,6c uma iqhathaniswa nomqaphi oseMhlabeni, kubonakala sengathi ingakanani kumuntu oqaphayo oseMhlabeni?
Ingxoxo:
Ukufinyela kobude kungenye imiphumela yokuhlobana echazwe ngokuhlobana okukhethekile, okuvezwa yi:
\[ L = L_0 \sqrt{1 – \left(\frac{v}{c}\right)^2} \]
lapho \( L_0 \) bubude bento ephumulile, \( v \) buyijubane elihlobene, kanye \( L \) buyibude bento ejubane elihlobene. Ngendiza:
\[ L_0 = 100 \umbhalo{amamitha}, \; v = 0,6c, \umbhalo{bese} \]
\[ L = L_0 \sqrt{1 – \left(\frac{v}{c}\right)^2} = 100 \sqrt{1 – (0,6)^2} = 100 \sqrt{1 – 0,36} = 100 \sqrt{0,64} = 100 \times 0,8 = 80 \text{amamitha} \]
Ngakho-ke, ubude bendiza ngokusho kwababukeli eMhlabeni bungamamitha angu-80.
Isibonelo Umbuzo 3: Amandla adonsela phansi kanye nethiyori yobudlelwano obujwayelekile ku-GPS
Umbuzo:
Amasathelayithi e-GPS azungeza uMhlaba endaweni ephakeme ngamakhilomitha angama-20.200 ngaphezu kobuso boMhlaba ngesivinini esingaba ngu-3,874 km/s. Usebenzisa ukuhlobana okuvamile, bala ukulungiswa kwesikhathi okudingeka kwenziwe amasathelayithi e-GPS nsuku zonke ukuze alandise imiphumela yamandla adonsela phansi oMhlaba.
Ingxoxo:
Amasathelayithi e-GPS kumele alungise isikhathi sawo ngenxa yemiphumela emibili eyinhloko: ukwanda kwesikhathi ngenxa yesivinini esiphezulu (ukuhlobana okukhethekile) kanye nokwanda kwesikhathi ngenxa yokudonsela phansi (ukuhlobana okuvamile). Kodwa-ke, sizogxila emphumeleni wokudonsela phansi lapha:
Kusetshenziswa inkolelo-mbono yokuhlobana okujwayelekile, isikhathi sizodlula kancane ensimini enamandla yamandla adonsela phansi. Ifomula yokuhlobana okubonakalayo okuvela ekuhlobaneni okujwayelekile yile:
\[ t_g = t_0 \kwesobunxele( 1 – \frac{2GM}{Rc^2} \kwesokudla) \]
lapho \( R \) kuyibanga elisuka enkabeni yamandla adonsela phansi, \( G \) kuyisimo esingaguquki, \( M \) iyisisindo soMhlaba, \( c \) iyisivinini sokukhanya, kanye \( t_0 \) yisikhathi sombukeli 'ongaguquki' ebusweni boMhlaba.
Kunikezwe:
– Ubuningi boMhlaba, \( M \cishe 5,972 \izikhathi eziyi-10^{24} \umbhalo{ kg} \)
– Irediyasi yoMhlaba, \( R_{\text{surface}} \cishe 6.371 \izikhathi 10^6 \text{m} \)
– Ukuphakama kwesathelayithi, \( H = 20.200 \izikhathi ezingu-10^3 \umbhalo{m} \)
– Ngakho ibanga elisuka enkabeni yoMhlaba liye esathelayithi, \( R = R_{\text{surface}} + H \approx 26.571 \times 10^6 \text{m} \)
Umehluko wesikhathi ngosuku phakathi kwesathelayithi kanye nobuso boMhlaba, kucatshangelwa amandla adonsela phansi kuphela:
\[ \Delta t_g \approx \frac{2GM}{c^2} \left( \frac{1}{R_{\text{surface}}} – \frac{1}{R} \right) \]
Ukuyishintsha:
\[ \Delta t_g \approx \frac{2 \times 6,67408 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2} \times 5,972 \times 10^{24} \text{ kg}}{(3 \times 10^8 \text{ m/s})^2} \left( \frac{1}{6,371 \times 10^6 \text{ m}} – \frac{1}{26,571 \times 10^6 \text{ m}} \right) \]
Ngemva kokubala, lo mphumela ulingana nokulungiswa kwesikhathi kwansuku zonke kwamasathelayithi e-GPS, okuhamba kancane ngama-microsecond angu-7 kunesikhathi sobuso boMhlaba. Ngakho-ke, amasathelayithi e-GPS kudingeka acabangele lo mphumela ukuze alondoloze ukunemba.
Umthelela Omkhulu Ebuchwephesheni Nokuqonda Indawo Yonke
Lezi zibonelo zikwenza kucace ukuthi ukuhlobana kuka-Einstein akuyona nje inkolelo-mbono engokoqobo kodwa futhi kunezindlela eziningi ezisebenzayo. Kusukela ekwandeni kwesikhathi ekuhambeni emkhathini kuya ekuncipheni kobude kanye nokulungiswa kwesikhathi kubuchwepheshe be-GPS, ukuhlobana kuka-Einstein kube nomthelela omkhulu.
Ukusungula izinto ezintsha emikhakheni eyahlukene yobuchwepheshe, isayensi, ngisho nefilosofi kubonisa ithonya lale nkolelo-mbono. Ukuhlobana kwezinto kuye kwavumela ukuhlola isikhala okujulile, ukuthuthukiswa kobuchwepheshe bokuxhumana obuthuthukisiwe, kanye nokuqonda okusha kwe-cosmology kanye nemigodi emnyama.
Ekugcineni, inkolelo-mbono ka-Einstein yokuhlobana kwezinto isalokhu iyingxenye ebalulekile yokutadisha i-physics yanamuhla futhi iyaqhubeka nokuba umthombo wokuphefumulelwa nokuhlola kososayensi emhlabeni jikelele.
