Question about rearranging formula in Brian Cox's Why Does E=MC2

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DPR

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This has been driving me insane. I don't get how he went from s/c to t/y. If someone could explain step by step how you do it I would greatly appreciate it.

Recall that we arrived at an expression for the length of the momentum vector in three-dimensional space, mΔx/Δt. We have just argued that Δx should be replaced by Δs and Δt should be replaced by Δs/c to form the four-dimensional momentum vector, which has a seemingly rather uninteresting length of mc. Indulge us for one more paragraph, and let us write the replacement for Δt, i.e., Δs/c, in full. Δs/c is equal to [sqrt (cΔt)^2)-(xΔ)^2]/c. This is a bit of a mouthful, but a little mathematical manipulation allows us to write it in a simpler form, i.e., it can also be written as Δt/γ where y=1/[sqrt 1-v^2/c^2)]. To obtain that, we have used the fact that υ = Δx/Δt is the speed of the object. Now γ is none other than the quantity we met in Chapter 3 that quantifies the amount by which time slows down from the point of view of someone observing a clock fly past at speed.
pg 127
 

DPR

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could someone help me out with this or give me some tips on what to study to be able to figure this out?
 

Matterwave

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s is defined as [itex]s=\sqrt{-(\Delta x)^2+c^2(\Delta t)^2}[/itex]

Now divide both sides by [itex]c\Delta t[/itex] and you get:

[tex]\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}[/tex]

Now notice that [itex]\frac{\Delta x}{\Delta t}=v[/itex] to get:

[tex]\frac{s}{c\Delta t}=\sqrt{-\frac{v^2}{c^2}+1}[/tex]

Move the t to the right hand side and use the definition of gamma to get:
[tex]\frac{s}{c}=\frac{\Delta t}{\gamma}[/tex]
 

DPR

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hey thanks! I really need to brush up on my math!
 

Matterwave

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One additional point. We often call [itex]\frac{s}{c}[/itex] the proper time [itex]\tau=\frac{s}{c}[/itex]. You may see this proper time pop up more often depending on which sources you're reading.
 

DPR

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One additional point. We often call [itex]\frac{s}{c}[/itex] the proper time [itex]\tau=\frac{s}{c}[/itex]. You may see this proper time pop up more often depending on which sources you're reading.
Thanks. How exactly did you go do this step? Now divide both sides by [itex]c\Delta t[/itex] and you get:

[tex]\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}[/tex]

I get everything after that.
 
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Thanks. How exactly did you go do this step? Now divide both sides by [itex]c\Delta t[/itex] and you get:

[tex]\frac{s}{c\Delta t}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}[/tex]

I get everything after that.
[tex]\frac{s}{c\Delta t}=\frac{\sqrt{-(\Delta x)^2+c^2(\Delta t)^2}}{\sqrt{(c\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2+c^2(\Delta t)^2}{c^2(\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+\frac{c^2(\Delta t)^2}{c^2(\Delta t)^2}}=\sqrt{\frac{-(\Delta x)^2}{c^2(\Delta t)^2}+1}[/tex]
 

Matterwave

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Just divide both sides like I said. You have to move the [itex]c\Delta t[/itex] inside the squareroot and divide both sides by it.

EDIT: What elfmotat said.
 

DPR

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thanks guys!
 

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