# md65536

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7

1. ## A grid and a shape

We're trying to prove 2 opposite things. Does trying to prove that an arbitrary shape can be placed without touching a point, lead to a contradiction?
2. ## The sum of the series

How did you solve it? I noticed a pattern in what each subsequent term was "leaving out", and figured out a formula for the sum of the first n terms, then used induction to see that it works as n goes to infinity. Is there another way?
3. ## ... paradox help needed ...

I concur. This sort of thing has cropped up before, and it has always been due to relativity of simultaneity (usually). Forget about the fluid for now. Consider the frame where both Vs are traveling toward each other at the same speed. They're length-contracted to an identical shape, and should therefore collide at all points along the V simultaneously. Therefore neither is there an enclosed volume to trap fluid nor extra space for the fluid to escape. The same events happen in all frames, ie. all points along the Vs must collide. The main difference is that the events aren't all simultaneous in other frames. Yes, in one frame the point will contact first, and in another the ends collide first. When you add the fluid back, in all frames the outcome is the same: either the fluid is moved early enough so it can escape in all frames, or it doesn't have enough time to move in any frame. You can maintain a paradox by stipulating some impossible requirement. Eg. have the 2 Vs come to mutual rest instantly in multiple frames. Eg. accelerate a perfectly rigid body (including a volume of incompressible liquid).
4. ## Momentum and spacetime curvature

I think that applies to all elementary particles in Newtonian gravity. I think I've been tricked into showing that I was wrong all along. Even with a Newtonian approximation of GR, adding energy somewhere does not necessarily simply add a Newtonian gravitational effect there. Even when the added energy adds rest mass to a system, it might not behave as if the mass was added where the energy was. Back to the original problem of 2 masses, it's possible it works approximately well when the masses pass, ie. with the system treated as a particle that gains rest mass as measured from a distance. However, when adding energy to each of the 2 masses, the behavior is not the same as if you increased their individual rest masses. To try to get away from the Newtonian view, mass is not some kind of "stuff" that exists independently, it's just the measurement. Of course I'd rather have simple answers but I suppose that's best, especially when bad intuitive explanations can stick around for decades, being repeated in pop sci media etc.

6. ## Momentum and spacetime curvature

I think my main confusion was taking this to mean that adding "mass" energy generally doesn't correspond with any "stronger" Newtonian gravity. I'm also not interested in Newtonian gravity except as an approximation of the predictions of GR. But as you pointed out, in my example the strong frame-dragging effect has no approximation in Newtonian gravity. It should be possible to contrive an example where the Newtonian forces are small, and the purely GR aspects dominate and give results that are the exact opposite of what Newtonian gravity predicts. Maybe, but in this example mass alone and the notion of energy having mass in the COM frame seems to work fine and fits with intuition. With 2 beams in the same direction, there's no COM frame. The energy of the beams is frame-dependent, and as total momentum approaches zero, so does the energy. Intuitively there is no energy that could make up "rest mass." With beams in opposite directions there is energy in the COM frame. True, that's not enough to guess how much they'd attract. I wouldn't even be certain the beams would attract (because of the uncommon geometry of the "mass"). As you've said such a notion of mass would be "problematic" in GR. But I think at least it's intuitive that the energy that doesn't vanish with a choice of coordinates, curves spacetime around the beams, similar to how any other equivalent mass would (and I don't know how to say that without using the word 'mass' even if it's just in a vague way).

8. ## Momentum and spacetime curvature

Then what is the correct way to describe their difference? Say you have 2 metrics, one for a system with little mass, and another for one with greater mass. The one with greater mass fits all intuitive notions of "stronger gravity", but how would that be expressed correctly? Greater mass corresponds with greater curvature, wouldn't the latter have greater curvature?
9. ## Momentum and spacetime curvature

I don't see how that is relevant so I wonder if we're even talking about the same thing? Where is my mistake in the following? All else being equal, a rotating object will have a higher invariant mass than the same object when not rotating. Curvature is related to invariant mass, in particular a higher invariant mass corresponds with a stronger Newtonian gravitational field. An object made of two massive bodies moving around each other, will have different spacetime curvature and "stronger" Newtonian gravity compared to the same bodies not moving (all else being equal). This answer's OP's main question with "yes." This also seems to contradict your statements. As far as I can tell, you're saying that because you can't plug the relevant values directly into the field equations, the result will be more complicated? But that won't change the statements above. That answer's OP's last question with "yes". I don't see how that matters. If you change the conditions that give you a different mass or different motion, you get different curvature, a different metric. Why would that imply anything about combining separate metrics?
10. ## Momentum and spacetime curvature

Yes, the rest masses don't change, but together the rest mass of the system includes the kinetic energy of the bodies in the system's COM frame. For example, the rest mass of an atom includes the (small) contribution of the kinetic energy of its electrons. This additional mass affects both the curvature around the atom, and the atom's Newtonian gravitational strength. No one's talking about doing that. I think a more appropriate analogy is comparing the Schwarzschild metric and the Kerr. If a non-rotating spherically symmetric mass (not necessarily a BH) is rotated, what changes? Its shape, and kinetic energy? But spacetime curvature is different, and so is its mass. Can you still say that the motion of the mass's components doesn't contribute to its mass or gravitational influence? It's similar for non-symmetric systems, such as 2 masses orbiting each other. Its mass and gravitational influence can include more than just the rest mass of its components. It's true for 2 orbiting bodies or for an atom or anything else that has a COM frame.
11. ## Momentum and spacetime curvature

I figure it is in this case. (There was another topic about this...) If you have 2 masses moving relative to each other, then there should be a center-of-momentum frame for them. In that frame, the kinetic energy of the masses contributes to the total energy of the system, and its rest mass. It should contribute to gravity just like any mass would? In the case where the two masses are at relative rest with each other, they're at rest in the center-of-momentum frame, so the system has no kinetic energy to contribute to its rest mass.
12. ## Recording or perceiving the activity of an oncoming object

They use "apparently faster-than-light motion" to mean the same as "apparently faster than c", so my earlier distinction between the 2 was pointless. I wish they'd distinguish though, because "faster than light" can mean "faster than light measured correctly (measured differently than the object)" or it can mean "can beat light in a race (measured the same way as the object)", and they mean only the former. Then people get in arguments and it turns out they're saying the same thing but meaning 2 different things. If an object can appear to move faster than c, then light can appear to move faster than c, and "light can appear to move faster than light" is confusing unless it's clear that it's implying 2 different measurements.
13. ## Recording or perceiving the activity of an oncoming object

Mostly no. Any light inside the event horizon will "fall" toward the singularity, but not all at the same rate (light sent inward will get there faster than light sent outward). No light from inside can go outside to the spaceship. Or, light from the spaceship can't go to places that light going past it can't... if we can't see what's in the black hole then either the spaceship can't see it or we can't see the spaceship. The event horizon passes over the ship at the speed of light, locally. It can "see" light that's part of the horizon, but it can't send signals faster than the light on the horizon. However once the spaceship enters the horizon, it could see some light from objects from inside the BH that fell in before it. Basically, it can fall toward the singularity more quickly than light directed outward does. Or, everything including light is moving toward the singularity, but you can catch up to light that is aimed toward you.
14. ## Recording or perceiving the activity of an oncoming object

Yes, that's correct. The relativistic Doppler effect includes delay of light and SR's time dilation. The Doppler factor period_received/period_sent = sqrt((1+beta)/(1-beta)) where beta*c is velocity, and for beta=-.6 the factor is equal to 0.5. https://en.wikipedia.org/wiki/Relativistic_Doppler_effect Yes it appears as if it's moving faster than c, but not faster than light, because incoming light appears to arrive at the same time it appears that it was sent! To look faster than that (v < -c), you'd have to see it arrive before you saw it leave, see it in 2 places at once, etc. The incoming (outgoing) object would also appear stretched (compressed) by the relativistic Doppler factor, so if something could approach the speed of light you could see it arrive just after it left, appearing stretched along its entire path, as well as bright and "very" blue. Sure, like if you consider an object falling into a black hole and sending out signals. A signal sent outward as the event horizon passes, becomes a part of the event horizon. Then if another observer falls in, it could see that signal as the event horizon passes it. I think that if both observers fell in at similar speeds, a photon sent exactly at the event horizon's passing wouldn't be redshifted, because where would the energy go? (It's not like the horizon itself stretches out.) I may be wrong. It seems, if you were both far above a black hole, and one went in, the other would see the one in front slow and redshift to the point of appearing stopped and invisibly dark. Then if the other followed, they'd see the one in front redshift less and less until appearing normal speed at the event horizon, and I think they'd then see the one in front blueshift increasingly redshift again and eventually "end" as there's some point where any "outward"-directed light from after that point will reach the singularity before you can catch up to it. That is, if the interior of the black hole has Schwarzschild geometry. (I think it would be blueshifted because the light you could receive would be increasingly later in the first object's life, it would have to appear sped up???) Edit: On second thought it seems that light would get "stretched" in opposite directions on either side of the event horizon, and it would immediately increasingly redshift as soon as the event horizon passed you. However, I'm trying to figure it out intuitively and could be way off. This is related to a topic that came up before, and I found it useful to see the paths of light on a Kruskalâ€“Szekeres diagram, when talking about what infalling observers might see. https://en.wikipedia.org/wiki/Kruskalâ€“Szekeres_coordinates Edit2: Okay looking at those diagrams again, the paths of light are 45-degree diagonal lines, now I remember what the diagrams showed: The second infalling observer could see images of the first observer from when they were "above" where the 2nd observer is now (higher r value). The 2nd observer would hit the singularity while still being able to "see" the first, but could only see the 1st's life up to some minimum r, whose value depends on how long the 2nd went in after the 1st. That implies the second observer would see the 1st redshifted inside the black hole.
15. ## Lorentz transformations btw rotating frames of reference

The transformations are between the coordinates in the respective reference frames, and I think you can/must choose the coordinates for Earth here. For a spherically symmetric rotating Earth with nothing else in the universe, the Kerr metric should work (an exact solution), and Kerrâ€“Schild coordinates is apparently a good choice. I think this would be coordinates where the Earth is rotating, not a rotating frame where Earth is completely stationary, but I suppose an additional transformation between those could be used. But I think you could choose different metrics, including ones without exact solutions, and different coordinate systems that I think could include or not some of the negligible relativistic effects. For the transformation, would you basically derive the Lorentz transformation using Kerrâ€“Schild coordinates?
16. ## The twin Paradox revisited

Yes, it seems that was the only thing missing? Don't worry about errors, that's the point of the thread, where something appears paradoxical when presented in a way that each part of it makes sense, but some missing thing is hidden. You were right---at least if we're talking about events only---that if (event) A is in the middle of (events) B and C in its own frame, it's in the middle in every frame. I was wrong, it doesn't matter where where the origin of the frame is. The Lorentz transformation is a linear transformation, so it preserves the 4D linearity of those 3 events (preserving the midpoint), but not distance between them or simultaneity.
17. ## The twin Paradox revisited

Yes, it's not correct for anybody. What I wrote earlier makes no sense, I'll try again... You have frames A, B, and C coincident at time t=0. In A's frame, B and C are equidistant from A. Let's call events BA and CA the events on B's and C's respective world lines that are at time t=0 in A's frame. Then you transformed BA and CA into B's frame and found that they're also equidistant from the event "A at proper time 0" in B's frame because of the way you put B's origin at that event. So far so good. In B's frame, at time t'=0 where you've set it up so that A is at the origin at that time, neither events BA nor CA have coordinate time t'=0. You've set it up so that they are away from the origin (x' != 0), so if you transform their time t into t' you should find that those events are not simultaneous in this frame. Therefore it's not the distances between the objects A, B, and C that you're comparing, but distances to events at different times. I guess the basic idea is that the Lorentz transformation you used applies to events, not the objects. (Note that the location of B is fixed in B's frame, so the distance to the origin is always the same and t' doesn't really matter for measuring the distance to B---I messed that up in an earlier reply---but the distance to C does depend on what time t' you measure it.)
18. ## The twin Paradox revisited

Easy for you, maybe... with your own assumptions. But it's a good demonstration of the point of the thread's topic. You can apply SR and use its equations, then slip one little detail from one frame into to another inappropriately, and you come up with a different answer.
19. ## The twin Paradox revisited

Yes, I made a mistake but revised it. You didn't show that the distance from A "to B" is the same as between A and C, in B's frame. You showed it for the distance from A "to the location of B at time t=0 as measured in A's frame", which is not the location of B in B's frame at time t'=0, because you have B not at the origin of B's frame. I really think you've overcomplicated it.
20. ## The twin Paradox revisited

I still think you should start with what you understand and build up from there, but... One thing I see is that you have A at the origin, and are looking only at time t=0 where the origins of A's frame and B's coincide. Yes, distances from the origin should scale by gamma. However, that stops working for t != 0, when the 2 frames' origins no longer coincide. Another way to look at it is that in B's frame, the velocity of C is not 2 times the velocity of A, so A does not maintain its place in the middle of B and C. Also, to get things to align at this one time, you've set up everything really unconventionally. You have B not at the origin of its own frame. I think you're transforming "the starting location of B, in A's frame" into a coordinate in B's frame... but B is not actually at that coordinate at time t'=0 in B's frame! Since you have B far from the origin in its own frame, the time at that location will be transformed. So basically you're saying "In B's frame, A is in the middle of where C will be at some time in the past or future, and where B will be some time in the past or future," which I think is correct. This is too complex and confusing for me.

22. ## The twin Paradox revisited

Careful, this can cause confusion. C has at least 2 reference frames. The frame you're calling "C's reference frame" is the frame where C is on the return journey to the meeting point. If that was clear to everyone in the discussion, that's fine. Edit: Nevermind me, I missed where the discussion switched to a case with no acceleration. I guess that's fine because you're basically looking at only the last phase of a twin paradox experiment. But in the full twin paradox experiment... in the C_outbound frame, B's clock is slower and B does everything that C does, later than C does. If they have a gradual acceleration phase, then when C is momentarily at rest with the "meeting point frame", B's clock has caught up and they've reached their maximum distance at exactly the same time in this frame. From then on, according to C, B's clock is ahead and anything that happens to C has happened earlier to B. So, in the case of only the inbound B and C frames, and the midpoint frame where B and C are symmetrical and always the same age, it is relativity of simultaneity that makes them not the same age in other frames (except at the meeting event). In B's frame before meeting, C is older but ageing slower.

24. ## The twin Paradox revisited

Yes, and different people are arguing different things. Some are saying only that if one twin has a certain velocity and then a different velocity, there "needs" to be coordinate acceleration for that to be possible. Others will say that acceleration is the "cause" of the difference in proper time. Some say that swapping the twin with another moving in the opposite direction is not "the Twin paradox" because it describes a different experiment even if the results are the same. Others argue any number of things (they can't be meaningfully swapped, the accelerated twin physically changes, the results would be different---I can't think of any examples I think are correct). "Feeling" the acceleration or having proper acceleration shouldn't matter because it's not generally in the calculations, and the time dilation effect still happens in freefall, eg. with GPS satellites. Well that's not true. It's valid in a single reference frame (the aging of both twins can be figured out using only the inertial twin's reference frame even if the other one is never inertial) and it's valid for comparing any number of inertial reference frames (eg. composition of velocities, using triplets all leaving together and returning again, etc.). The first video says something like "SR only applies to uniform motion", and that's wrong. It generally applies only in uniformly moving reference frames, and I think that's implied by its postulates, which include that the speed of light is a constant c, and it's not constant in an accelerating reference frame. However, SR does apply to non-uniform motion of objects. For example, if the accelerating twin spends the entire trip accelerating, SR can be used to calculate its time dilation from an inertial reference frame just fine, just from its velocity.
25. ## Can a material object cross the event horizon of the Black Hole?

Yes of course it's possible and common. Google says 'About 1% of supermassive black holes have an "accretion disk" of gas and dust swirling around them.' So I guess it looks like an object being pulled apart from the inside. If you want to consider some other object, it should be similar. Some "forces" give the object its shape and keep parts of the object away from the black hole, whether it's the inertia of a spinning accretion disk or chemical bonds of solid matter, and the BH rips away any parts that get too close, regardless of what holds the object together. It would look like spaghettification. The process is just gravitation, not some magical suction that vacuums everything up. For example a human on an extremely massive planet would be crushed by gravity. A black hole with similar gravitation would devour a human from the inside because the latter doesn't have the structural integrity to resist such strong pull. A small enough asteroid would barely have the strength to keep a human stuck to it, let alone damage its shape. A black hole with similar gravitation wouldn't have the strength to rip a human apart or pull more of it in.
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