Misali na Tambayoyin Tattaunawa Game da Dangantaka
Dangantaka tana ɗaya daga cikin muhimman ra'ayoyi a fannin kimiyyar lissafi na zamani, wanda Albert Einstein ya gabatar a farkon ƙarni na 20. Wannan labarin zai tattauna ka'idar dangantaka da kuma yadda take aiki a rayuwar yau da kullum ta hanyar misalai da bayanai.
Gabatarwa ga Dangantaka
Ka'idar dangantaka ta ƙunshi manyan sassa biyu: Ka'idar dangantaka ta musamman da kuma Ka'idar dangantaka ta gaba ɗaya. Ka'idar dangantaka ta musamman, wacce aka buga a shekarar 1905, ta kawo sauyi ga fahimtarmu game da sarari da lokaci. A cikin wannan ka'idar, Einstein ya bayyana cewa saurin haske shine iyakar gudu mafi girma da ba za a iya wuce shi ba kuma dokokin kimiyyar lissafi iri ɗaya ne ga duk masu lura da ke tafiya a cikin gudu mai ɗorewa.
A halin yanzu, Ka'idar Dangantaka ta Gabaɗaya, wacce aka gabatar a shekarar 1915, tana magana ne game da nauyi. A ƙarƙashin wannan ka'idar, nauyi ba ƙarfin gargajiya bane, amma lanƙwasa ne na lokacin sarari da lokaci wanda taro ke haifarwa.
Fahimtar wannan ra'ayi na asali yana da matuƙar muhimmanci kafin mu shiga cikin tambayoyin misalai da tattaunawarsu.
Tambayoyi da Tattaunawa Samfura
Tambaya ta 1: Faɗaɗa Lokaci
Tambaya:
Wani ɗan sama jannati yana tafiya zuwa wani tauraro mai nisa a gudun 0,8c (inda c shine saurin haske). Idan tafiyar ta ɗauki shekaru 10 a duniya, nawa ne ɗan sama jannatin zai fuskanta bisa ga agogonsa (lokacin da ya dace)?
Tattaunawa:
Faɗaɗa lokaci wani abu ne da ke faruwa saboda bambancin saurin da ke tsakanin masu kallo biyu. Lokaci yana wucewa a hankali ga abu yana motsawa idan aka kwatanta da mai kallo da ke tsaye.
Tsarin faɗaɗa lokaci shine:
\[ \Delta t' = \frac{\Delta t}{\sqrt{1 – \frac{v^2}{c^2}}}\]
Ina:
– \(\Delta t'\) shine lokacin da aka lura da abin da ke motsi.
– \(\Delta t\) shine lokacin da aka lura da wani abu da ba ya tsayawa.
– \(v\) shine saurin abin da ke motsi.
– \(c\) shine saurin haske.
Sanya dabi'un da aka sani a cikin dabarar:
\[ v = 0,8c \]
\[ \Delta t = 10 \, \rubutu{shekara} \]
\[ \Delta t' = \frac{10}{\sqrt{1 – \frac{(0,8c)^2}{c^2}}}\]
\[ \Delta t' = \frac{10}{\sqrt{1 – 0,64}}\]
\[ \Delta t' = \frac{10}{\sqrt{0,36}}\]
\[ \Delta t' = \frac{10}{0,6}\]
\[ \Delta t' \approx 16.67 \, \text{year}\]
Don haka, lokacin da ɗan sama jannatin ya samu a daidai lokacin da yake da shi, kimanin shekaru 16,67 ne.
Tambaya ta 2: Matsewar Tsawon Lokaci
Tambaya:
Tsawon abu yana da mita 100 kuma ana auna shi a lokacin hutawa. Idan abin yana motsi a gudun 0,6c, nawa ne tsawon abin bisa ga mai lura da shi?
Tattaunawa:
Takaitaccen tsayi wani abu ne da ke faruwa inda tsawon abin da ke motsawa idan aka kwatanta da mai lura ya fi guntu fiye da lokacin da abin yake hutawa.
Tsarin da ake amfani da shi wajen rage tsawon lokaci shine:
\[ L = L_0 \sqrt{1 – \frac{v^2}{c^2}} \]
Ina:
– \(L\) shine tsawon abin da ke motsi.
– \(L_0\) shine tsayin da ya dace (tsawon abin idan yana hutawa).
– \(v\) shine saurin abu.
– \(c\) shine saurin haske.
Sanya dabi'un da aka sani a cikin dabarar:
\[ L_0 = 100 \, \rubutu{mita} \]
\[ v = 0,6c \]
\[ L = 100 \sqrt{1 – \frac{(0,6c)^2}{c^2}}\]
\[ L = 100 \sqrt{1 - 0,36}\]
\[ L = 100 \sqrt{0,64}\]
\[ L = 100 \sau 0,8\]
\[ L = 80 \, \text{mita}\]
Don haka, tsawon abin da ke motsi bisa ga mai lura da shi a tsaye shine mita 80.
Tambaya ta 3: Tsarin Alaƙa
Tambaya:
Kwayar cuta tana da nauyin hutu na kilogiram 2. Idan wannan ƙwayar cuta tana motsi a gudun 0,9c, menene nauyin ƙwayar cuta?
Tattaunawa:
Tsarin Relativistic shine nauyin abu wanda ke ƙaruwa yayin da abu ke matsawa kusa da saurin haske.
Tsarin ma'aunin relativistic shine:
\[ m = \frac{m_0}{\sqrt{1 – \frac{v^2}{c^2}}} \]
Ina:
– \(m\) shine jimlar nauyin da ke nuna alaƙa.
– \(m_0\) shine sauran taro (daidaitaccen taro).
– \(v\) shine saurin abu.
– \(c\) shine saurin haske.
Sanya dabi'un da aka sani a cikin dabarar:
\[ m_0 = 2 \, \rubutu{kg} \]
\[ v = 0,9c \]
\[ m = \frac{2}{\sqrt{1 – \frac{(0,9c)^2}{c^2}}}\]
\[ m = \frac{2}{\sqrt{1 – 0,81}}\]
\[ m = \frac{2}{\sqrt{0,19}}\]
\[ m \approx \frac{2}{0,436}\]
\[ m \kimanin 4,59 \, \text{kg}\]
Don haka, nauyin da ke tattare da ƙwayar cuta lokacin da take motsawa a gudun 0,9c shine kimanin kilogiram 4,59.
Tambaya ta 4: E=mc^2
Tambaya:
Nawa ne makamashin da ake samarwa idan gram 1 na wani abu ya lalace gaba ɗaya bisa ga dabarar Einstein \(E=mc^2\)?
Tattaunawa:
Shahararren dabarar Einstein \(E=mc^2\) tana ba da alaƙa kai tsaye tsakanin taro (m) da kuzari (E), tare da \(c\) shine saurin haske.
A cikin tsarin SI (Tsarin Ƙasashen Duniya na Ƙungiyoyi):
– Ana auna nauyi (m) a cikin kilogiram (kg).
– Gudun haske (c) shine \(3 \sau 10^8 \, \text{m/s}\).
Bari mu ƙididdige kuzarin da aka samu daga gram 1 na wani abu:
– gram 1 = 0,001 kg
\[ E = mc^2 \]
\[ E = (0,001) (sau 3 10^8)^2 \]
\[ E = (0,001) (sau 9 10^{16}) \]
\[ E = 9 \sau 10^{13} \, \text{joules} \]
Don haka, kuzarin da ake samarwa idan gram 1 na abu ya lalace gaba ɗaya shine joules (sau 9 10^{13}\).
Kammalawa
Dangantaka muhimmin ra'ayi ne a fannin kimiyyar lissafi, tare da zurfafan ma'anoni ga nau'ikan abubuwan da suka shafi zahiri. Ta hanyar misalan da aka tattauna a sama, mun ga yadda za a iya amfani da ka'idar musamman ta dangantaka don fahimtar fadada lokaci, takaita tsawon lokaci, yawan dangantaka, da kuma dangantakar da ke tsakanin taro da makamashi.
Ta hanyar fahimtar da kuma aiwatar da waɗannan matsalolin, za mu iya fahimtar kyawun ka'idar dangantaka da kuma tasirinta ga fahimtar sararin samaniya.
