What am I doing wrong while proving time dilation using Minkowski space-time diagram?
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Hey mate ^_^
It's not important at all, but just so you know, the "official" symbol you're looking for is not "\triangle" but "\Delta". Doesn't matter for this post (the difference between △△ and ΔΔ - and we all know what you mean), but if you ever use LaTeX to type up a paper to be published, a PI might care
#Be Brainly ❤️
It's not important at all, but just so you know, the "official" symbol you're looking for is not "\triangle" but "\Delta". Doesn't matter for this post (the difference between △△ and ΔΔ - and we all know what you mean), but if you ever use LaTeX to type up a paper to be published, a PI might care
#Be Brainly ❤️
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Hello mate here is your answer.
So, just to clarify your approach:
You take events A and C occurring at the same location x′a=x′cxa′=xc′ but different times t′ata′ and t′ctc′ in the primed frame. You also take event B occurring at the same time as A in the primed frame, t′b=t′atb′=ta′, but at the same location as C in the unprimed frame, xb=xcxb=xc. For events B and C you then apply the invariance of the space-time interval
c2(tb−tc)2−(xb−xc)2=c2(t′b−t′c)2−(x′b−x′c)2c2(tb−tc)2−(xb−xc)2=c2(tb′−tc′)2−(xb′−xc′)2
and in view of xb=xcxb=xc, t′c=t′atc′=ta′, obtain
c2(tb−tc)2=c2(t′a−t′c)2−(x′b−x′c)2c2(Δt)2=c2(Δτ)2−(Δx′)2c2(tb−tc)2=c2(ta′−tc′)2−(xb′−xc′)2c2(Δt)2=c2(Δτ)2−(Δx′)2
Lastly, the Lorentz transform of Δx′=x′b−x′cΔx′=xb′−xc′ gives Δx′=γ(xb−vtb)−γ(xc−vtc)≡γvΔtΔx′=γ(xb−vtb)−γ(xc−vtc)≡γvΔt, and you conclude, correctly, that cΔt=cΔτ/γcΔt=cΔτ/γ.
Your "problem" is that regardless of the substitution t′c=t′atc′=ta′ your final relation still gives (tb−tc)=γ(t′b−t′c)(tb−tc)=γ(tb′−tc′). Let's tally up what we have:
If we use events B and C in the unprimed frame, but A and C in the primed frame, we find that."The time interval between events B and C occurring at the same location in the unprimed frame appears time dilated wrt the time interval between event C and an event A occurring in the primed frame at the same location as C but at the same time as B".The last piece of information, "occurring in the primed frame at the same time as B", is the crucial one: we can replace event A with any other event, at any location, as long as it "occurs in the primed frame at the same time as B".Otherwise, if we dispense with event A and simply refer to events B and C only, we just find that.The time interval (tb−tc)(tb−tc)between two events B and C occurring at the same location in the unprimed frame appears time dilated in the primed frame.
Hope it helps you.
So, just to clarify your approach:
You take events A and C occurring at the same location x′a=x′cxa′=xc′ but different times t′ata′ and t′ctc′ in the primed frame. You also take event B occurring at the same time as A in the primed frame, t′b=t′atb′=ta′, but at the same location as C in the unprimed frame, xb=xcxb=xc. For events B and C you then apply the invariance of the space-time interval
c2(tb−tc)2−(xb−xc)2=c2(t′b−t′c)2−(x′b−x′c)2c2(tb−tc)2−(xb−xc)2=c2(tb′−tc′)2−(xb′−xc′)2
and in view of xb=xcxb=xc, t′c=t′atc′=ta′, obtain
c2(tb−tc)2=c2(t′a−t′c)2−(x′b−x′c)2c2(Δt)2=c2(Δτ)2−(Δx′)2c2(tb−tc)2=c2(ta′−tc′)2−(xb′−xc′)2c2(Δt)2=c2(Δτ)2−(Δx′)2
Lastly, the Lorentz transform of Δx′=x′b−x′cΔx′=xb′−xc′ gives Δx′=γ(xb−vtb)−γ(xc−vtc)≡γvΔtΔx′=γ(xb−vtb)−γ(xc−vtc)≡γvΔt, and you conclude, correctly, that cΔt=cΔτ/γcΔt=cΔτ/γ.
Your "problem" is that regardless of the substitution t′c=t′atc′=ta′ your final relation still gives (tb−tc)=γ(t′b−t′c)(tb−tc)=γ(tb′−tc′). Let's tally up what we have:
If we use events B and C in the unprimed frame, but A and C in the primed frame, we find that."The time interval between events B and C occurring at the same location in the unprimed frame appears time dilated wrt the time interval between event C and an event A occurring in the primed frame at the same location as C but at the same time as B".The last piece of information, "occurring in the primed frame at the same time as B", is the crucial one: we can replace event A with any other event, at any location, as long as it "occurs in the primed frame at the same time as B".Otherwise, if we dispense with event A and simply refer to events B and C only, we just find that.The time interval (tb−tc)(tb−tc)between two events B and C occurring at the same location in the unprimed frame appears time dilated in the primed frame.
Hope it helps you.
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