Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Feb 8, 2012 #1

    DPR

    User Avatar

    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.

     
  2. jcsd
  3. Feb 16, 2012 #2

    DPR

    User Avatar

    could someone help me out with this or give me some tips on what to study to be able to figure this out?
     
  4. Feb 16, 2012 #3

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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]
     
  5. Feb 16, 2012 #4

    DPR

    User Avatar

    hey thanks! I really need to brush up on my math!
     
  6. Feb 16, 2012 #5

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  7. Feb 16, 2012 #6

    DPR

    User Avatar

    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.
     
  8. Feb 16, 2012 #7
    [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]
     
  9. Feb 16, 2012 #8

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  10. Feb 16, 2012 #9

    DPR

    User Avatar

    thanks guys!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about rearranging formula in Brian Cox's Why Does E=MC2
  1. Why E=MC2? (Replies: 5)

  2. Confused about e=mc2 (Replies: 3)

  3. In E=mc2, Why C? (Replies: 57)

  4. Questions about E=mc^2 (Replies: 25)

Loading...