Small problem with getting to E=mc^2

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  • #1
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First of all, I'm not sure where to post this.

Now, I am reading some lectures about relativity, I came quite far and understand the concept. The steps in the following I just can't follow.



"Kinetic Energy and Mass for Particles of Arbitrary Speed

For a particle of rest mass m0 accelerating along a straight line (from rest) under a constant force F, "

[PLAIN]http://img840.imageshack.us/img840/8366/relapic.jpg [Broken]

The first step I understand, but then I'm completely lost. Spend over 2 hours in figuring it out...
I know I'd have to use m= m0 divided by root of (1-v2/c2)

Can someone help me?
 
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Answers and Replies

  • #2
DrGreg
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Does this clue help?

[tex]\frac{dm}{dt} = \frac{dm}{dv}\,\frac{dv}{dt}[/tex]​
 
  • #3
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well, it does explain where that dv/dt comes from..

But I still have problems,

Now I am thinking like this: ( which might be completely wrong)

dm/dv multiplied with v = what is on the left of the + sign without the dv/dt

and dm = m - m0
and dv must be something like v -v0 which means dv=v (right?) and dt therefore = t ?
dv must be written in another way.. but what?

so.. still stuck. thanks for the first clue though :smile:
Would you be so kind to give an other clue?

p.s. going to study math and physics next year in Utrecht university (Netherlands) so my skills aren't at a really high level..
 
  • #4
Doc Al
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Realize that m = γm0. (Which you note in your first post.) So dm/dv = m0 dγ/dv.
 
  • #5
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Well, I think I'm almost there...
I just can't figure out what to make of the dv (and looking ahead, I probably won't be able to do the next step either, but that depends on the solution =/ )
Thanks alot.
 
  • #6
DrGreg
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Well, I think I'm almost there...
I just can't figure out what to make of the dv (and looking ahead, I probably won't be able to do the next step either, but that depends on the solution =/ )
Thanks alot.
Well, I assumed from your first post that you knew about calculus and that "dy/dx" does not mean dy divided by dx, but means the derivative of y with respect to x. If you don't know that, then none of this will make much sense and you'll need to study differential calculus first.

If you do know calculus, then use the fact that

[tex]\frac{d}{dv}\,\frac{1}{\sqrt{z}} = \left( \frac{d}{dz} \, \frac{1}{\sqrt{z}} \right)\frac{dz}{dv}[/tex]​
 
  • #7
jtbell
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That is, use the chain rule for derivatives.
 
  • #8
Fredrik
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I did a similar calculation here. Maybe that will help. Note that I'm using units such that c=1, and the symbol m for what you call m0.
 
  • #9
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I got it now =D I was looking totally in the wrong direction.
[tex]\frac{d}{dv}\,\frac{1}{\sqrt{z}} = \left( \frac{d}{dz} \, \frac{1}{\sqrt{z}} \right)\frac{dz}{dv}[/tex]​
z=1-v2/c2
Y=1/sqrt(z)

first the dm/dt becomes dm/dv * dv/dt
dm/dv becomes m0 *dY/dz *dz/dv
and then first part.
dz-0,5/dz the derivative is
-0.5z-1,5

now we have dz/dv left which is quite easy
-2v/c2

so, now multiplying (1st deriv = -2v/c2) * (2nd deriv = -0.5z-1,5) *dv/dt * v * m0
-2 and -0,5 multiplied is 1 so crossing that off and we're there.

THANKS FOR ALL YOUR HELP
 
  • #10
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But I'm not there yet. the last step is also a bit too much for me.

help?
 
  • #11
Fredrik
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If you mean the last equality in the image you posted, this is the key part of it

[tex]\frac{v^2}{1-v^2}+1=\frac{v^2+1-v^2}{1-v^2}=\gamma^2[/tex]

This is with c=1. I highly recommend using units such that c=1. If you want to keep that c around, just replace the v in the denominator with v/c, and choose between doing the same thing in the numerator or inserting a factor of c2 next to the gamma on the right. (When you restore factors of c, you just insert them where they're needed to ensure that the things we're adding have the same units, and that both sides of each equality have the same units).

Edit: I should also have mentioned this: You know that when you have an expression of the form A+B with B≠0, you can rewrite it as B(A/B+1). This trick, with "B" equal to the entire second term in the expression you have to simplify, simplifies that expression considerably. Then you can just do what I did above.
 
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  • #12
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Thanks, I think I got it.

Love you all =D
 

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