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Relativistic forces

  1. Apr 22, 2013 #1
    Am working my way through relativistic mechanics, and one force transform (Fx) I derive has an extra factor in one term. Of the many textbooks I have, the only one that presents those equations is Resnick – Introduction to Special Relativity. And it doesn’t look like even he derived them, but copied them from another ~1960 reference that I can’t find. So it is possible that there could be a typo mistake. Can anyone confirm his results for that transform? I’ll try to type his equation for the x-force (along the common x-axes, prime system has positive relative velocity, etc.)

    Fx’ = [Fx – beta u*F/c]/[1-ux v/c^2]

    My extra factor is 1/[1-u*u/c^2] in the (beta u*F/c) term.

    (u and F are vectors, * is the dot product, v is relative frame velocity along the x-axes, u is the object velocity, ux is the object velocity component parallel to the x-axes, etc.)
  2. jcsd
  3. Apr 22, 2013 #2


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    No typo. Check the case where u and F are both in the x-direction. Then from the previous equation, Fx = Fx', and his result agrees with that without any add'l factors.

    Resnick is a really poor book, IMO. He invariably makes things 10 times as complicated as they need to be. My advice is to sell it on eBay and focus on the other books instead.
  4. Apr 22, 2013 #3

    Meir Achuz

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    A gamma in the numerator cancels one in the denominator. You might have used 1/gamma in the denominator by mistake. The derivation just takes dp'/dt' and Lorentz transforms the dp' and the dt' to dp and dt.
    It only takes two steps.
  5. Apr 23, 2013 #4
    I agree about Resnick, but his text is the only one I have that even gives this formula (and I have quite a few). Do you have some other recommendation? And I must be missing something else obvious: do we know that F'x = Fx is correct for that situation?

    Yes, that's what I have been trying to do. Could you show me the 2 steps? And note that my extra term is not gamma. It looks like gamma but with the object's velocity in the un-primed system rather than the relative velocity between reference frames.
  6. Apr 24, 2013 #5

    Meir Achuz

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    =\frac{dp/dt-v(dE/dt)}{1-vdx/dt}=\frac{F-v(u\cdot F)}{1-uv},[/tex]
    with c=1.
  7. Apr 26, 2013 #6
    Now I need to go back and see if I can derive that momentum-energy transform.
  8. Apr 29, 2013 #7
    Question on geometric nature of general relativity

    Here's my question. I am googling "forceless" and "inertia" in connection with general relativity. I have read, not so much Einstein's followers, but Einstein's publications. I never saw where he actually said that general relativity was forceless. Here is my problem. Mach was big, big on inertia, whereby masses in the distant cosmos have an effect on local physical laws. Does not that mean that masses near each other will increase? I know that Einstein said exactly that. Namely, that inertia increases when other masses are pilled up in its "neighborhood". I have found only one source by googling on line that remotely even talks about such increased inertia in this scenario, and it's totally bespoke (a third new science). Are any of you accomplished enough in solving the Einstein Equation to say whether this "relativistic" mass notion is fact or fiction? If masses do increase when stars / planets converge, then why is their gravitational potential energy lower when they converge? This is what cobb points out on the site as a "paradox". But it seems to make at least some sense. Help ! Why is not the dark matter problem at least coinsidered in light of this type of mass increase when billions of stars are in close proximity to each other? I just going to sign up to get the free "treatise" in any event. He having a public experiment. Maybe it will flop

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