Pure SR, Relativistic Mass, and its Gravity?

Discussion in 'Physics & Math' started by Neddy Bate, Jun 5, 2018.

  1. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    Q-reeus, the electric field example is different from the rocket case Neddy and I have been discussing. In the electric field example, you have a constant force applied along the y-axis, but the acceleration is not going to be strictly along the y-axis. In the rocket example, you have a constant acceleration along the y-axis (for some finite period of time), but a non-zero force along the x-axis, which nonetheless cancels off when considering equal and opposing forces applied to the propellant in order to conserve total x-momentum.

    Also you make some errors in your electric field calculations. Kinetic energy is given by \((\gamma-1)mc^2\), not \(\gamma mc^2\). I believe the correct final angle should be \(\arctan(\sqrt{5/3})\).
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. Q-reeus Valued Senior Member

    Messages:
    3,003
    And? What do you think my #154 was sorting out for all to ponder?
    In other words you now concede I was correct on that. It's as discussed in my last post, and of course in #154.
    Again, yes - as given in #154.
    Should I take that last bit as a typo? Surely you mean constant \(v_x\)? Otherwise you appear to be contradicting your (now also mine) position baseball momentum along x-axis increases when tossed in that train. Which by analogy will likewise apply to the rocket scenario. If you just mean there is am overall conservation of momentum, well nobody here is disputing Newton's 3rd Law. Of course there is also an overall conservation of total y-momentum.

    Anyway the situation with rocket can be viewed as the propellant shooting out at other than strictly along the y-axis when there is velocity both along x and y axes in a given inertial frame K. What that angled thrust seen in K is cancelling to maintain constant \(v_x\) is the otherwise kinematic loss of velocity that a purely y-acting force e.g. the qE of Case 1 in #154, would ensure.
    Yes I made an error there. Was up all night trying to nut out the issue and got quite tired by the time I posted. It doesn't propagate to change the rest of it anyway.
    No, for Case 1 with values as specified, the final angle is 45 degrees. \(p_x\) and \(p_y\) have equal magnitude seen in K frame. To repeat, gain in \(p_y\) owing to action of qE there has no effect on \(p_x\), unlike the case in baseball-train or rocket scenarios.
     
    Last edited: Jun 21, 2018
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    No, I'm pointing out that your electric charge example doesn't justify the errors you continue to make in analyzing the rocket scenario, you're comparing two fundamentally different scenarios.

    No, it's not a typo. I'm pointing out that the rocket as seen in the K frame is able to maintain a constant x-velocity even though its x-momentum is not constant, and that there's nothing screwy about this, because the net x-momentum of rocket + propellant is still conserved. A vertical force seen in the K' frame does not correspond to a vertical force seen in the K frame, although the acceleration in both frames is strictly directed along their y-axes.

    Actually, the error you made there does indeed propagate through to the rest of the problem. You shouldn't be up all night trying to figure out how to cover some very simple mistakes, instead of just owning up to them and apologizing for berating others who didn't make the same mistakes. The particle starts off with rest mass mc^2 and kinetic energy mc^2, moving along x. A force along the y-axis then causes this kinetic energy to double, so the particle now has a total energy of 3mc^2. In this situation, the final momenta along x and y are not equal.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Q-reeus Valued Senior Member

    Messages:
    3,003
    That was all done and shown in #154. With admittedly a few errors in maths that do not effect the fundamental physics concepts discussed. Clearly you can't let it rest there.
    Again, nobody but you keeps bringing in expelled propellant. As per #110, the only interest is in a constant force accelerating a specified *payload* along a given axis as seen in some inertial frame K', and how that is effected by moving to a different frame K.
    Yes and no. In #154 it was initially specified that KE gain owing to qE acting over time along y = initial coasting KE along x. So the final KE should have shown as
    \((\gamma -1)mc^2\) = \((3-1)mc^2\) = \(2mc^2\).
    With final \(\gamma = 3\) there, not the 4 shown. Upon substituting that value of 3 back in
    v = \(\sqrt(1-\frac{1}{\gamma^2})\)c = \(\sqrt(1-\frac{1}{3^2})\)c = \(\sqrt(1-\frac{1}{9})\)c = 0.94281c
    Final trajectory is at 45 degrees since I had initially specified that KE gain owing to motion along y = initial KE along x. So the \(v_x\), \(v_x\) components have equal magnitude given by \(\frac{v}{\sqrt{2}}\) = 0.6667c. Which, again, has final \(v_x\) obviously less than initial value 0.866c. That general finding remains true.
    Thus, contrary to your assertion in #157, specifically directed at the Case 1 scenario of #154:

    "If a particle is traveling at 0.866c along the x-axis, you can't accelerate it strictly along the y-axis to reach a 45-degree angle. The particle's inertial mass would approach infinity as it approached lightspeed, and the Pythagorean velocity sum would always remain below c with the x-velocity remaining constant. If you were to instead boost the lab frame along y without affecting the particle, the x-velocity component would now be reduced in this boosted frame and you could pick a frame in which the particle is now traveling at 45 degrees. You can't boost the observer's reference frame when you want to talk about conserved quantities, you have to pick a fixed inertial frame and leave it in place."

    Which is basic conceptual error. You should have picked me up on the incompatibility of my initial specification requiring \(p_x\) = \(p_y\), with the incorrect final value for \(\gamma\) I gave in #154. Instead, you ignored the initial specification, and made a big thing about my mistaken value for final \(\gamma\).
    See above. Maybe you should finally own up to your basic conceptual error in #157. Or do you wish to continue maintaining, that the charged particle \(v_x\) is an invariant in K frame of #154? Thus continue claiming that generation of \(p_y\) owing to a strictly y-axis qE force will somehow boost \(p_x\) so as to maintain \(v_x\) constant? It won't. Case 1 & Case 2 are fundamentally different. You agreed to that much in #161 but your #157 actually denies it.
     
    Last edited: Jun 22, 2018
  8. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    That's still wrong. Even though the same amount of kinetic energy is transferred by forces along each axis, that doesn't mean equal momenta have been transferred. Change in energy is the integral of force over distance, change in momentum is the integral of force over time- not the same thing. Your reasoning would be correct if we were working with Newtonian mechanics, but we're not. \(\arctan(\sqrt{5/3})\). Stop arguing and do the calculation on paper, as I did.
     
  9. Q-reeus Valued Senior Member

    Messages:
    3,003
    So show your paper calculations here, allowing the error to be pinpointed. For sure, my specification re KE's guarantees \(p_x\) = \(p_y\) in magnitude. Thus also velocities.
    OK on a rethink I had that wrong. My bad. Magnitude of \(p_y\) will be less than \(p_x\), owing to having to accelerate along y an increased transverse inertial mass.
    So one would need a larger increase in KE to have \(p_x\) = \(p_y\).
    Ironically, it would have to be twice the initial KE, thus my initial mistaken value \(\gamma\) = 4 is then the right one to have \(p_x\) = \(p_y\).
    It's still always possible to achieve 45 degrees. It's still true boosting \(p_y\) via qE has no effect on \(p_x\)
    And btw, you haven't dealt with my pointing out your conceptually wrong #157.
     
    Last edited: Jun 22, 2018
  10. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    I never disputed that the electric charge example can result in a 45 degree deflection, but not with the numbers you were working with when I last did the calculation. As to #157, nothing is conceptually wrong with it. If, with the numbers given, an acceleration (NOT force!!!) is applied strictly along the y-axis, it cannot lead to a 45 degree deflection, regardless of how long a constant vertical acceleration is maintained in the rocket frame.
     
  11. Q-reeus Valued Senior Member

    Messages:
    3,003
    Sorry but that won't fly. It's disingenuous to talk apples when the subject is oranges. Your #157 is clearly meant to dispute my claim in #154 a force qE strictly along y-axis can easily yield a final trajectory at 45 degrees, or any other value less then 90 degrees from x-axis. And this will have no effect on \(p_x\). I never once there even hinted let alone specified acceleration had to be strictly along y-axis in K frame, for Case 1 scenario your #157 specifically addressed. Given the x-axis velocity decreases, always my point, obviously acceleration in K has to have an x-axis component! And for good measure the distinctions between charge in E field vs rocket etc. was summarized at bottom of #154.

    I got stuck in edit mode too long doing a rework in #166, so it all went poof. Here's what that (streamlined) rework yields, as to required gain in KE owing to impressed qE along y, to give 45 degrees final trajectory.

    Magnitude of initial coasting \(p_x\) = \(\gamma_x mv_x\) = 2(0.866c)m = 1.732mc.
    Which \(p_x\) value doesn't change regardless of |\(p_y\)| = |\(p_x\)| being attained owing to action of qE force along y. Net momentum in K is simply
    |p| = |\(\sqrt{2}p_x\)| = 2.449mc. From e.g. https://en.wikipedia.org/wiki/Energy–momentum_relation,
    \(\gamma = \sqrt{1 + (\frac{p}{mc})^2}\)
    Substituting |p| = 2.449mc, gives \(\gamma = \sqrt{1 + (2.449)^2}\) = \(\sqrt{6.997}\) = 2.645
    Of which the net KE contribution is \(\gamma -1\) = 1.645. The additional KE owing to action of qE over some time interval \(\Delta_t\) is thus
    \((1.645 - 1)mc^2\) = \(0.645mc^2\)

    It actually requires less added KE than that initially contained in coasting along x-axis, to achieve a 45 degree trajectory.
    Which on reflection makes sense as more time \(\Delta_t\) is required, hence greater impulse qE\(\Delta_t\) to go from \(v_y = 0\) to \(v_y = |v_x|\),
    owing to the transverse inertial mass \(\gamma_x m\) exceeding proper mass m were \(v_x = 0\) in K frame.
     
    Last edited: Jun 22, 2018
  12. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    Now I'm starting to repeat myself, the clock is ticking on how many times I'm going to do this. Once again, as I have been saying EVERY SINGLE TIME, the rocket example and your charge example are DIFFERENT. In the charge example, I myself agreed that the x-velocity changes, and I calculated that the deflection angle will be even greater than 45 degrees, specifically \(\arctan(\sqrt{5/3})\).

    In the case of the rocket, there is NO horizontal acceleration and NO change in the initial horizontal velocity, as seen in either reference frame. There is no rocket veering, there is no failed launch as long as the thrust is vertical in the rocket's rest frame. If the rocket accelerates with a constant thrust along the y-axis, as seen in the rocket's instantaneous frame of reference, then:

    1) The horizontal velocity as seen by the stationary ground observer is \(v_x\) and NEVER, EVER, EVER changes by even so much as a scant little nanometre/second. The acceleration is purely along the y-axis in either frame.
    2) The rocket never reaches the speed of light. As the speed of light is approached, the vertical acceleration falls off as seen by the ground observer, assuming it's constant in the rocket's frame.
    3) As a corollary to 2), if the rocket is moving faster than \(\frac{c}{\sqrt{2}}\) along the x-axis, it never reaches a 45 degree angle.
    4) As seen by the static ground observer, the horizontally gliding rocket will accelerate purely vertically with \(v_x\) held constant, but will appear to experience a horizontal force leading to an exchange of x-momenta between the rocket and its expelled propellant.

    Capisce?
     
  13. Q-reeus Valued Senior Member

    Messages:
    3,003
    That was all explained and agreed upon in #154. Gone over again in #160. There were errors in certain specific values earlier on there but now cleaned up in #168. Capisce? Why repeat it now in BOLD? It doesn't begin to explain your #157, which interestingly enough, totally avoided the #154 Case 2 section and summary. I guess you wished to give a certain negative impression. For instance, last part in #157:
    "You can't boost the observer's reference frame when you want to talk about conserved quantities, you have to pick a fixed inertial frame and leave it in place."
    Again, disingenuous. Suggesting I had not done so when it's perfectly clear Case 1 in #154 was worked out (with subsequently acknowledged numerical errors, all fixed in #168) in the one K inertial frame. The conserved quantity being \(p_x\) in K frame.
    Yes, somewhat in #159, and more clearly #161, there was express agreement on my points re fundamental differences, raised in #154, then #160. Not before. Not in #157.
    See above.
    As I mentioned in an earlier post, there is though an increasing tilt seen in the K frame when both x & y axes velocities are non-zero. Similarly for the exhaust initial direction. Lorentz contraction.
    1), 2), & 3) are all covered and agreed upon in #154, with 2) & 3) above conforming to that given in last line there. No need to repeat again in BOLD.
    See above. See #154, #160.
    You too, I hope. Maybe time to move on.
     
    Last edited: Jun 22, 2018
  14. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    There's no basis to move on to anything if, after all this, you still can't grasp how an acceleration applied strictly along one axis leaves the velocity constant along any other axis, even though the accompanying forces can be off-axis. The rocket does not skew, it does not tilt.
     
  15. Q-reeus Valued Senior Member

    Messages:
    3,003
    Can't grasp? Really, you do have this nasty streak. And yes the rocket does tilt, increasingly as y speed builds, seen in K frame where the rocket has non-zero x-axis speed. Lorentz contraction.
     
  16. Neddy Bate Valued Senior Member

    Messages:
    1,586
    I admit I didn't read through #154 because it seemed to be about capacitors, charges, or whatever. Now that I realize there was also an agreement to CptBork's 1), 2), and 3), I went back and dug it up. I am glad we are all finally in agreement on the x-component velocity remaining constant for the original rocket scenario. My thanks to CptBork for bringing Q-reeus around. I don't think my arguments were going to cut it alone.
     
  17. Q-reeus Valued Senior Member

    Messages:
    3,003
    So is it all done to your satisfaction? Apart from the x-velocity thing that is - what of the original query about g-forces experienced? All settled?
     
  18. QuarkHead Remedial Math Student Valued Senior Member

    Messages:
    1,550
    Not to mine! Ya know, the Captain is quite right - given a minimal set of rectilinear coordinates, then nothing that "happens" exclusively relative to one coordinate can possibly have any effect on anything that "happens" relative exclusively to another (independent) coordinate.

    Basic stuff....
     
  19. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Messages:
    5,682
    Yeah I mean if you have a rocket launching off a platform and there's a massive tower connected to the platform which goes straight up the y-axis for 50 lightyears, and someone's moving along the x-axis at a constant velocity relative to that platform, they're not going to see the rocket veering off course from the tower, and they're not going to see the tower bending over to the side.

    I've tried several times to explain, there's a difference between a constant acceleration along y with varying forces along x, and a constant force along y with varying accelerations along x. Forces and accelerations don't always have to be parallel in Relativity.
     
  20. Neddy Bate Valued Senior Member

    Messages:
    1,586
    I think we can forget all about the x-velocity thing for now.

    Let's focus on the simple laboratory frame accelerating up some y axis with the result being a constant "1 gee" experienced inside the lab. Also inside the lab, we have the treadmill, moving at constant horizontal speed, as measured by the lab.

    From the above scenario, I had offered two possible ways to measure the 'weight' of a test mass:

    1. We can place a spring scale underneath the entire treadmill, zero it out to neglect the treadmill's own weight, and then place a test mass on the belt to measure the test mass weight.
    2. We can place a spring scale on top of the treadmill belt, with a test mass on top of the scale, and measure the test mass weight.

    1. If I understand correctly, #1 results in the test mass weighing exactly like it would on earth, except multiplied by a factor of gamma.
    2. If I understand correctly, #2 results in the test mass weighing exactly like it would on earth, except multiplied by a factor of gamma squared.

    1. Can anyone please confirm, #1 has the scale itself experiencing '1 gee' because the scale is stationary with respect to the lab, and therefore calculates the mass of the test mass to be its 'relativistic mass' because it is increased by a factor of gamma?
    2. Can anyone please confirm, #2 has the scale itself experiencing 'gamma squared gees' because the scale is stationary with respect to the top belt of the treadmill, and therefore calculates the mass of the test mass to be its 'rest mass' which would make sense since the scale is stationary with respect to the test mass?

    Then we can finally be done here. Thank you.
     
  21. Q-reeus Valued Senior Member

    Messages:
    3,003
    At this stage, you will probably want confirmation of above 1. & 2. (correct as per my #13), by going to an authority elsewhere. I pointed to one such back in p3, #56, #57. Go check out those posts linked to over there again, and see if anything further is required.
     
  22. Confused2 Registered Senior Member

    Messages:
    496
    Probably too simple but I'll have a go anyway...
    In the lab frame anything on the belt moves up by y=(1/2)gt²
    In the belt frame we have t'=γt
    Since y is perpendicular to the x velocity y'=y
    so
    (1/2)gt²=(1/2)g'(γt)²
    so g=(γ)²g'
     
  23. Q-reeus Valued Senior Member

    Messages:
    3,003
    Implicitly the t's there ate actually delta t's i.e time intervals not moments in time. It works out ok with that approach, except you should have the last line reading
    g' = (γ)²g
    , given g' > g is the required belt frame value, and g the given one in lab frame.
    Also, it would be a good idea to specify you are using the 'dropped object' approach there.
     
    Last edited: Jun 26, 2018
    Confused2 likes this.

Share This Page