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Relativistic Energy- well a basic algebraic simplification

  • Thread starter Sneil
  • Start date
  • #1
18
0
a little rough on this simplification, is this correct?
solving for "u"

E = mc^2 / (root)(1-u^2/c^2)

(root)(1-u^2/c^2) = mc^2/E

1 - u^2/c^2 = (mc^2/E)^2

1 = (mc^2/E)^2 + u^2/c^2

c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)


u = c - mc^3/E

pretty bad i don't know this basic simplification, but it'll all come back quickly enuf

thanks for the help :smile:

-Neil
 
Last edited:

Answers and Replies

  • #2
120
0
c^2 = (mc^2/E)^2 (c^2) + u^2 -> (sq root everything)


u = c - mc^3/E

Your mistake occurs on the transition between these two lines

when you take the root of a function which has two terms added together, it is not equivalent to the root of each one added together

ie root (a^2 + b^2) is not equal to (a + b)

so you should really have
u = root ((mc^2/E)^2(c^2) + u^2)
 
  • #3
18
0
alright thank you :smile:


Warr said:
so you should really have
u = root ((mc^2/E)^2(c^2) + u^2)
but do you mean

u = root ((mc^2/E)^2(c^2) + c^2) but actually u = root (c^2 - (mc^2/E)^2(c^2) ) :confused:
 
  • #4
120
0
ah sorry, yes thats what I meant to type

also you could further make it so that

u = root (c^2(1 - (mc^2/E)^2))
u = c*root(1 - (mc^2/E)^2)
 
  • #5
18
0
great! thanks a lot for the help :smile:
-Neil
 

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