Tambayoyi da Tattaunawa Kan Alaƙar Einstein
Dangantakar Einstein tana ɗaya daga cikin manyan ka'idoji a fannin kimiyyar lissafi na zamani, tana canza yadda muke fahimtar sarari da lokaci. Ta ƙunshi sassa biyu: Dangantakar musamman (1905) da Dangantakar gabaɗaya (1915). A cikin wannan labarin, za mu tattauna misalai da dama da suka shafi Dangantakar Einstein kuma mu tattauna su don samar da fahimta mai zurfi.
Dangantaka ta Musamman
Dangantaka ta musamman tana hulɗa da abubuwa masu motsi a saurin da ke kusantar saurin haske. Sakamako biyu masu mahimmanci na wannan ka'ida sune faɗaɗa lokaci da kuma rage tsawonsa.
1. Faɗaɗa Lokaci
Idan akwai masu kallo guda biyu, ɗaya a tsaye a Duniya ɗayan kuma yana tafiya da sauri sosai, za su auna lokaci daban-daban don wannan lamari.
Misalin matsalar:
Ɗan sama jannati yana tafiya sau 0.8 na gudun haske (c) zuwa ga tauraro mai shekaru 10 na haske daga Duniya. Tsawon wane lokaci ɗan sama jannati zai ɗauka kafin ya isa ga tauraro?
Tattaunawa:
Da farko, muna ƙididdige lokacin da mai lura a Duniya ya auna:
\[ t_B = \frac{d}{v} = \frac{10 \text{ shekaru masu haske}}{0.8 \, c} = 12.5 \text{ shekaru} \]
Don ƙididdige lokacin da ɗan sama jannatin ya auna (faɗaɗa lokaci), muna amfani da dabarar:
\[ t_A = t_B \sqrt{1 – \frac{v^2}{c^2}} \]
Maye gurbin dabi'un da aka sani:
\[ t_A = 12.5 \sqrt{1 – (0.8)^2} \]
\[ t_A = 12.5 \sqrt{1 – 0.64} \]
\[ t_A = 12.5 \sqrt{0.36} \]
\[ t_A = 12.5 \sau 0.6 \]
\[ t_A = 7.5 \rubutu{ shekaru} \]
Don haka, lokacin da 'yan sama jannatin suka auna shine shekaru 7.5.
2. Dogayen Matsewa
Idan abu ya motsa a gudun da ya kusanci saurin haske, tsawonsa zai yi kama da gajere ga mai lura da shi.
Misalin matsalar:
Jirgin sama mai tsawon mita 10 yana tafiya sau 0.9 fiye da gudun haske. Tsawon lokacin da jirgin zai ɗauka ga mai kallo a duniya?
Tattaunawa:
Don ƙididdige matsewar tsayi, muna amfani da dabarar:
\[ L = L_0 \sqrt{1 – \frac{v^2}{c^2}} \]
Ina:
– \( L_0 \) shine tsawon da ya dace ko tsawon gaske (mita 10),
– \( v \) shine saurin jirgin sama (0.9c).
Maye gurbin dabi'un da aka sani:
\[ L = 10 \sqrt{1 - (0.9)^2} \]
\[ L = 10 \sqrt{1 - 0.81} \]
\[ L = 10 \sqrt{0.19} \]
\[ L = 10 \sau 0.436 \]
\[ L = 4.36 \rubutu{ mita} \]
Don haka, tsawon jirgin a cewar masu lura da shi a Duniya shine mita 4.36.
Dangantakar Jama'a
Jumla ta gabaɗaya tana magana ne game da nauyi, inda sarari da lokaci ke tasiri ta hanyar taro da kuzari.
3. Ruwan tabarau mai nauyi
Ruwan tabarau na nauyi yana faruwa ne lokacin da haske daga wani abu mai nisa ya karkata ta hanyar nauyin wani babban abu kamar galaxy ko black hole.
Misalin matsalar:
Galaxy A tana da isasshen nauyi don karkatar da haske daga quasar B, wanda ke bayanta. Idan kusurwar karkatarwa ta kasance daƙiƙa 1.5 na baka, menene nauyin galaxy A? (Yi amfani da ma'aunin nauyi na Newton G = 6.674×10^-11 N(m/kg)^2, saurin haske c = 3×10^8 m/s)
Tattaunawa:
Ana iya bayar da kusurwar karkacewa θ ta hanyar dabarar:
\[ \theta = \frac{4GM}{c^2 R} \]
Ina:
– \( G \) shine ma'aunin nauyi,
– \( M \) shine taro na taurarin,
– \( c \) shine saurin haske,
– \( R \) ita ce tazara mafi kusa tsakanin haske da tsakiyar taurarin.
Tunda muna son nemo M, muna sake tsara dabarar:
\[ M = \frac{\theta c^2 R}{4G} \]
A ɗauka cewa R mita 5×10^20 ne (matsakaicin nisan taurari). Canza θ daga arcseconds zuwa radians (arcsecond 1 = 4.848×10^-6 radians):
\[ \theta = 1.5 \sau 4.848 \sau 10^{-6} \, \text{radian} = 7.272 \sau 10^{-6} \, \text{radian} \]
Maye gurbin dabi'un da aka sani:
\[ M = \frac{(7.272 \sau 10^{-6}) (sau 3 10^8)^2 (sau 5 10^{20})}{4 \sau 6.674 \sau 10^{-11}} \]
\[ M = \frac{(7.272 \sau 10^{-6}) (sau 9 10^{16}) (sau 5 10^{20})}{26.696 \sau 10^{-11}} \]
\[ M = \frac{(3.2764 \sau 10^{31})}{26.696 \sau 10^{-11}} \]
\[ M = 1.227 \sau 10^{41} \, \text{kg} \]
Don haka, nauyin tauraron dan adam na A yana da kimanin kilogiram 1.227×10^41.
4. Faduwar Perihelion na Mercury
Jumla ta gaba ɗaya kuma na iya bayyana yanayin sararin samaniyar duniyar Mercury wanda ba za a iya bayyana shi ta hanyar masana kimiyyar Newtonian ba.
Misalin matsalar:
Menene girman canjin perihelion na Mercury kamar yadda aka bayyana ta hanyar haɗin gwiwa gabaɗaya? (Sigar dangantaka A: 43 arcseconds a kowace ƙarni)
Tattaunawa:
Yi amfani da bayanan da aka bayar kai tsaye:
A cewar ka'idar Einstein ta gabaɗaya game da alaƙa, canjin perihelion na Mercury da aka bayyana shine daƙiƙa 43 arc a kowace ƙarni, wanda kuma yayi daidai da sakamakon lura.
Ƙarshe:
Ta hanyar kammala waɗannan misalan matsaloli da tattaunawa, za mu iya ganin yadda dangantakar Einstein ke ba da zurfin fahimtar lokaci, tsayi, da nauyi. Wannan ka'ida ba wai kawai ta canza ra'ayinmu na kimiyya game da sararin samaniya ba, har ma tana da aikace-aikace masu amfani a cikin fasahar zamani, kamar tsarin kewayawa na GPS, waɗanda ke buƙatar gyare-gyare na dangantaka don yin aiki daidai. Koyo da fahimtar dangantakar Einstein muhimmin mataki ne na zurfafa zurfafa cikin duniyar kimiyyar lissafi mai rikitarwa.
