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Thread: Types of mass

  1. #1
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    Types of mass

    I was talking to my physics professor a few years ago (wow, I guess it was 4 years ago). We got talking about mass. He mentioned there are really two types of mass. Relativity is based on the assumption that the two types of mass are really the same thing, and to the limits of our measurement techniques they have never been observed to differ. However, there is no actual reason to assume they are really the same, and more accurate measurements could conceivably show them to be different.

    The problem arises in that I do not remember what the two types are. I think one is the quantity that allows someone to compute change in velocity from the applied force or vice versus, although I am not sure. The other might have been used in computing gravitational interactions, but I am not as sure about that one. Anyway, I was wondering if anyone knows what the two are, and if you could give me any more details regarding
    the framework behind this. I think I understand the idea, what we refer to as "mass" is really two quantities derived from different physical interactions and they may or may not be the same on a fundamental level. There may be more to it, though.

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    Yes, the two types of mass are inertial and gravitational. Inertial mass is a measure of inertial, or resistance to acceleration, while gravitational mass is the "source charge" of the gravitational field.

    Inertial mass is the "m" in Netwon's second law, F = ma.

    Gravitational mass is the "m" in Newton's law of gravity,

    F = -G m1*m2 / r^2, which looks just like Coulomb's Law save for the minus sign.

    Let's define a gravitational field just like we define an electric field, E, as the ratio of force per unit charge. That's just F/m for gravity, which has dimensions of acceleration, and so we'll call the gravitational field just 'g'.

    g = -GM/r^2, where big "M" is some source. Now, imagine a little test mass in that field. The force is m_g*g, where "m_g" is *gravitational mass* of the test charge, which is analogous to F = qE.

    So F = m_g *g.

    Now, to get the acceleration, we must divide by the *inertial mass* of the test charge, "m_i", so

    a = (m_g / m_i) *g.

    Now, you can see if the two are equivalent, then a = g, trivially obvious (at first blush).

    However, what if they were different? Well if all particles had the same ratio of m_g to m_i then a would just be a constant that was the same for all particles times g.

    We couldn't tell the difference. We think we are measuring 'g', but we are really measuring 'a', and that 'a' would be the same so all masses still fall at the same rate. You can see how we could wrap up the constant ratio into "G", the gravitational constant, and express the gravitational force law in terms of inertial mass.

    As long as the ratio is constant, we can never tell the difference. That strongly suggests, but does not prove the two are the same. I think this constant ratio can be put right into GR and it work the same, even though the gravitational field is more than simple inverse square.

    -Richard

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    And more little mind game: The equivalence principle, which in a nutshell says that m_i = m_g is based on the idea of not feeling a force under gravitational acceleration.

    When I was young, I used to make the argument that the fact we feel no force of acceleration proves the equivalence principle, that inertial mass is the same as gravitational mass. If they were different, we would feel a force equal to the difference between the two, something like (m_i - m_g)*g Think about that a while, and you'll see.

    But, as someone straightened me out about, this is not so. I feel a force because of the internal stresses that transmit the force from one part of my body to all the others. Standing on an accelerating platform in free space, the platform pushes against my feet, and my feet then have to transmit that force through my body. This stress is what I feel.

    However, the gravitational force would act on all bits of my mass directly, and no internal forces would have to transmitted. And that would apply to an accelerometer as well. It would register no force, no matter what the ratio of inertial to gravitational mass actually is.

    -Richard

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    Cool Eotvos experiment

    The Hungarian physicist, Eotvos, designed experiments to see if the two masses are the same. Results say yes to about one part in 10-8...so there's no definitive difference. Different experiments involved checking nuclei with high proton/neutron ratios....and high neutron/proton ratios....within the island of nuclear stability...still null.
    We await the results of MINOS, which will clue in the triad...the reaction of these same said nuclei to neutrino fluxes. Eotvos experiments intrinsically assume that during their runs the neutrino flux runs isotropic. The same effect that a Mossbauer detector monitors as a reactor goes back online after refueling...is applicable to an Eotvos setup...but Mossbauer's are more sensitive. Trinity. Pete

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    Also note that the way to prove they cannot have a fixed constant of proportionality between m_i and m_g is that if you ever thought you found one, you are using a bad gravitational constant G. Think about it. So there is simply no point in ever imagining that they have a fixed ratio between them. Alternatively, you can set it up so that they have a fixed non-unity ratio any time you want-- just change G in all the reference books, and you're done.

    As for a varying ratio, it is true that the Eotvos experiment strongly suggests they are exactly the same, but here is a place where theory can trump experiment, in some sense. No matter how precise your measurement, no experiment will ever show they are exactly equal (so we'll never really know), but Einstein's wonderful theory says that gravity is not really a force at all, it is really more like a g. Given that, it is we who are inventing m_g, and making it the same as m_i, when we assert that there is some imaginary force of gravity that obeys Newton's law: F = m_i * a = m_g * g. It's all in our construction of this force, there's no such thing as gravitational mass.

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    ???Acceleration = Gravity???

    Quote Originally Posted by publius
    ...Now, you can see if the two are equivalent, then a = g,...
    Although the equation of the effect of acceleration equates with the equation of the effect of gravity, acceleration and gravity are not equivalent.

    In time dilation, accelerations have no effect on clocks - to understand an acceleration it must be broken down into a series of infinitely small velocity segments through integration to calculate the time dilation that occurs because of the acceleration. So an acceleration is just a constantly changing velocity and it is the velocity that affects the clocks.

    Gravity, on the other hand, can not be broken down into velocities to calculate the time dilation of gravity. A stationary object under 1g has more time dilation than a stationary object under 0g - there is no movement involved.

    So somehow the force of gravity slows clocks whereas the force of an acceleration does not directly affect clocks.
    Last edited by Squashed; 2006-Jun-29 at 01:10 PM.

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    Quote Originally Posted by Squashed
    Although the equation of the effect of acceleration equates with the equation of the effect of gravity, acceleration and gravity are not equivalent.
    Google "the equivalence principle".

    Quote Originally Posted by Squashed
    So somehow the force of gravity slows clocks whereas the force of an acceleration does not directly affect clocks.
    Actually, they both slow clocks in the same way. What you are missing is that time dilation is not the only affect in relativity-- there are also clock synchronization effects. People are always forgetting this, including me. The velocity effect you mentioned is a local effect only-- if you have acceleration of a distant object, it will also desynchronize its clock from yours. This is the twin "paradox" in a nutshell.

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    Quote Originally Posted by Squashed
    In time dilation, accelerations have no effect on clocks - to understand an acceleration it must be broken down into a series of infinitely small velocity segments through integration to calculate the time dilation that occurs because of the acceleration. So an acceleration is just a constantly changing velocity and it is the velocity that affects the clocks.
    Only if you want to use special relativity to do the calculations. Of course, we usually do want to do that, because it's a lot easier than using general relativity!

    Quote Originally Posted by Squashed
    So somehow the force of gravity slows clocks whereas the force of an acceleration does not directly affect clocks.
    No, if you're in a sealed room and experience 1 g of acceleration, you'll discover that clocks that are higher in the room run more quickly than clocks that are lower in the room, by exactly the same amount that they would if you were in a sealed room at rest on the Earth's surface. From your sealed room, you wouldn't be able to tell the difference.* For the accelerating room, you could be outside in an inertial reference frame and show how the observed time dilation results from differences in velocity between points that are higher or lower in the room. But for the Earth, you could do your calculations in a free-falling frame and also show that these differences in clock rates are from differences in velocity, rather than due to any force of gravity.

    * Actually, there is a slight difference of course. That's because for the accelerating room, the gravitational force is constant, whereas for the Earth, the gravitational force gets weaker as you get farther away and points in a slightly different direction (toward the center of the Earth) as you move side to side. The equivalence principle really says that acceleration and gravity are the same in an infinitesimal region where the force can be considered constant, or for practical applications, where these effects will be small enough to be ignored for whatever it is you're doing.
    Conserve energy. Commute with the Hamiltonian.

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    That's What I Thought Until ...

    Quote Originally Posted by Ken G
    Actually, they both slow clocks in the same way...
    That is not the impression I got when I read the Wikipedia article about how the acceleration of the test particle in the Fermilab experiment had no effect on the time dilation calculations.

    Wikipedia Quote under this sub-title:

    Origin of the Paradox

    "... It is to be noted, however, that time dilation has no relationship whatsoever to the amount, direction, or duration of acceleration. Even the time dilation of particles in the Fermilab ring is determined only by their speed, in spite of the huge accelerations they undergo. ..."

    Quote Originally Posted by Ken G
    ...there are also clock synchronization effects. People are always forgetting this, including me. The velocity effect you mentioned is a local effect only-- if you have acceleration of a distant object, it will also desynchronize its clock from yours. This is the twin "paradox" in a nutshell.
    Explain the above statement - why would an acceleration "close by" not affect time dilation while an acceleration "far away" would.

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    Quote Originally Posted by Squashed
    That is not the impression I got when I read the Wikipedia article about how the acceleration of the test particle in the Fermilab experiment had no effect on the time dilation calculations.
    Sure, at Fermilab we're calculating the effects of a particle's motion from a (nearly) inertial reference frame, so we're just using special relativity. If we wanted to do some hideously complex calculating, we could instead work everything out from the point of view of the particle, using general relativity to treat it as though it were at rest under the influence of a suddenly appearing gravitational field. The math would be really complex, though.

    I know we keep coming back to this, and that it's hard to grasp. I'll try to summarize. With the introduction of general relativity, an inertial reference frame is equivalent to one that is freely falling in a gravitational field. Also, an accelerating reference frame is equivalent to a stationary frame in a suitable gravitational field. There are no local experiments that can distinguish between these pairs of cases. However, an inertial reference frame is not equivalent to a stationary frame in a gravitational field, nor is a freely falling reference frame equivalent to one which is accelerating in the absence of gravity.
    Conserve energy. Commute with the Hamiltonian.

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    Quote Originally Posted by Grey
    * Actually, there is a slight difference of course. That's because for the accelerating room, the gravitational force is constant, whereas for the Earth, the gravitational force gets weaker as you get farther away and points in a slightly different direction (toward the center of the Earth) as you move side to side. The equivalence principle really says that acceleration and gravity are the same in an infinitesimal region where the force can be considered constant, or for practical applications, where these effects will be small enough to be ignored for whatever it is you're doing.
    The above tiny sized quote explains how the equivalence principle can equate the two: gravity and accelerations.

    But there are fundamental differences between accelerations and gravity:

    Gravity - tends towards a point; whereas,
    Acceleration - tends towards a spatial plane.

    Gravity - affects every atom of a body equally; whereas,
    Acceleration - is transmitted from atom to atom through the body.

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    Squashed,

    All of this illustrates how tricky the Equivalance Principle can be, namely "who is really accelerating".

    Being "at rest" in accelerating frame is the same as being at rest in a gravitational field. A free falling frame is the same as inertial frame from the POV of an accelerating frame.

    When you are at rest in a gravitational field, say standing on the earth, the ground supplies a *real force* to keep you from falling, or free-fall accelerating. That real force is transmitted atom-to-atom from your feet up, just like force would be transmitted if you were standing on an accelerating platform in free space.

    That is the source of much confusion -- some get confused and think it the free-falling observer that is equated with the accelerating observer. Not so, it is just the opposite:

    Accelerating observer = stationary observer in gravitational field who is resisting gravity. Both must feel a real force. GR says both of these are the ones who are really accelerating, as both are deviating from the geodesics of inertial motion.

    Free falling observer = inertial observer in flat space time. Neither feel any force, as both are following geodesics, and both are unaccelerated from the POV of GR.

    So standing on the ground, it is as though the ground is accelerating up, pushing you along with it. Watching an object in free fall is just seeing an inertial object *appear* to accelerate because you are the one who is "really accelerating".

    This is the Equivalence Principle.


    -Richard

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    Squashed,

    And other thing. Gravitational time dialation is due to *differences in gravitational potential*, not to differences in 'g' or gravitational acceleration. Consider one of the super massive black holes at the centers of galaxies. There is so much mass there that the value of 'g' at the event horizon itself is very small, around 1 earth 'g'. However, time appears to stop at the event horizon for stationary observers, just like for any black hole.

    The reason is the enormous difference in gravitational potential. Classically that goes as -GM/r, g goes as -GM/r^2. Now, let M become astronomically large. The 1/r^2 function has a long tail where it gets close to zero. It doesn't change much way out in the tail, but M is so large is makes it "linger" around 1 g for a very long distance. A object free-falling from infinity would reach a classical velocity of c at this point, which is the basic definition of the event horizon. And the difference in potential is enormous as well. However, g remains small.

    In an accelerating frame, we can construct a similiar gravitational potential function and find the clock rates depend on that. There is even an event horizon for the accelerating observer as well.

    And actually, there is a gradient to this equivalent gravitational field as well for the accelerating observer. This is because of Lorentz contraction. Imagine a long rod being accelerated. From our inertial frame, that rod contracts as its velocity increases. But what does that means. Since the fore end is getting shorter, it doesn't have to accelerate as fast as the middle. The middle "gains on" on the fore end from our frame. And likewise the rear end has to accelerate faster since it is getting closer to the middle.

    So there must be a gradient in the 'g' the accelerating observer measures. And there is. But this field is still different from an inverse square field, but a gradient is there nonetheless.

    So from the POV of the accelerating observers, objects far away in the foreground (high up in the psuedo-gravity well) don't appear to be accelerating much. Looking in the rear view mirror, objects appear to accelerate faster and faster the farther they get until they dissappear behind the psuedo-event horizon. And yes, their clocks appear to stop when they cross that horizon.

    But from the POV of an inertial observer watching that whole show, things are very different and much more normal acting. We can calculate the clock rates and distances entirely by SR integrating the velocity changes.

    But from the frame of the accelerating observer, SR does not work, and we have to include the "psuedo-gravitational field" in addition to relative velocity to get relative clock rates and all that stuff.


    -Richard

    -Richard

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    Excellent Post

    Quote Originally Posted by publius
    ... So there must be a gradient in the 'g' the accelerating observer measures. And there is. But this field is still different from an inverse square field, but a gradient is there nonetheless. ...
    publius,

    I read your first post and then went for a walk to think about the equivalence of the two: acceleration and gravity.

    My mind turned to the idea that if the two are identical then 5 seconds in a 1g field would be equivalent to 5 seconds with an acceleration of 1g and so I was thinking that the general relativity formula for time dilation would equal the special relativity formula for those 5 seconds worth of acceleration - integrated to convert to velocity.

    But then I realized that velocity accumulates under acceleration and so after 1.0 year at 1g the velocity would attain the speed of light and so for a 75 year old person time would have to stop at some point in their life.

    Then I read your 2nd post.

    If the two are equivalent then shouldn't the gradient be the same for both: an inverse square relation?

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    Squashed,

    You'll just have to take my word that it's not inverse square, but I can't show it, as that calculation is a toughie. When I was a young buck and had this stuff fresh in my head, I may have tackled, but not now.

    What you would do is, say, imagine the long rod being accelerated and put the origin of a coordinate system right at the center. We would pin our "constant acceleration" value, just call it 'g0' at the center. Now, (all from an inertial frame watching the show), you require that the length of the rod remain constant in an instantaneous inertial frame centered on the rod at all times. You then factor the Lorentz contraction relation with this requirement and you'll find a gradient in the acceleration you see across the length of the rod. Fore end less than 'g0', rear end greater than 'g0'.

    And this "acceleration field" will not be inverse square with distance but something else. It will be similiar but not the same. Let the fore direction be positive, call it 'r'. You'll find that g(r) asympotically approaches zero as r goes to infinity. For negative r, aft end, g(r) increases. Note this g(r) is in the frame of the accelerating observer.

    That g(r) will not be inverse square, but it will pretty close on the tail, which means for low values of g0, acceleration, the psuedo-field is pretty close to that of one of the super massive black holes. And both are not much different than a truly uniform field, g(r) = g0 = constant.

    The higher the acceleration, the more is going to deviate and not fit with an inverse square curve pinned at g0 at some origin.

    Remember, the Equivalance Principle is about a "small local region". The value of your acceleration determines how small it has to be. It is usually phrased in terms of little deviation from a uniform gravitational field. However, as we see above, it really means deviation from the actual g(r) of the accelerating observer.


    -Richard

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    Quote Originally Posted by publius
    Gravitational time dialation is due to *differences in gravitational potential*, not to differences in 'g' or gravitational acceleration.
    Yes, and this is why when you are talking about the effects of acceleration on time, the distance away that the acceleration occurs is very important. This is because you multiply the acceleration times the distance to get the change in gravitational potential. If you try to get this effect from special relativity, all you need to do is include the non-local clock desynchronization effect I was talking about. So I think there are really three levels of relativity: local special relativity in an instantaneously inertial frame, generalized special relativity including both integrals over instantaneous inertial frames and nonlocal clock desynchronization effects, and finally full-blown general relativity, which is really a theory of gravitational tidal effects, i.e., a fully nonlocal theory.

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    Ken,

    That's a very good point. And funny you mention tidal effects, because that's a little conundrum with the "psuedo-field" of the accelerating frame. There is actually a gradient in that field, as I was rambling about to Squashed. And so we should have tidal forces, no?

    But wait, switch to an intertial observer watching that accelerating observer go by. He is completely inertial, feeling no forces whatsoever, and certainly would feel no tidal forces. Just because some other character is accelerating by him should not make him suddenly feel tidal forces.

    So there must be something else going on that would allow the accelerating observer to calculate the "free falling" observer is actually experiencing no tidal forces.

    I don't know how the math would play out, but I know it would have to be something. Our 'g' gradient is derived from the Lorentz contraction, so I imagine that the changing Lorentz contraction the accelerating observer saw would just cancel out the 'g' gradient (since it is the same function) and eliminate the tidal stress forces.

    An inverse square field would not work out that way. But still that same Lorentz contraction effect would be there, just not cancel out completely somewho. So this leads me to think that the actual tidal forces a free falling body experiences will be different than the classical 1/r^3relation.

    Stationary (co accelerating) observers however, would "feel the gradient" across their length. It may be the tidal forces are actually velocity dependent. That is, it would be different at the same stationary point in the gravitational field if I dropped an object from different altitudes.

    This does get complicated.

    -Richard

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    Questions From the Audience

    Quote Originally Posted by publius
    But wait, switch to an intertial observer watching that accelerating observer go by. He is completely inertial, feeling no forces whatsoever, and certainly would feel no tidal forces. Just because some other character is accelerating by him should not make him suddenly feel tidal forces.
    The above brings to mind interesting questions (to me, at least): Does the spectator appear to be length contracted to the accelerated participant?

    Does the spectator appear to be length contracted to the participant in an inertial state traveling at a high velocity?

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    Inertial States

    Quote Originally Posted by publius
    ... Free falling observer = inertial observer in flat space time. Neither feel any force, as both are following geodesics, and both are unaccelerated from the POV of GR. ...
    This part is a bit confusing. Einstein claimed that a freefalling body was an inertial body (as does the above statement) and so I made the claim that an orbiting satellite was inertial (or near inertial) because it was in continual freefall and yet others claim that it is not inertial because it is in a rotating frame which requires continual change in direction and so it is not inertial.

    publius, do you have any words to clarify my confused state?

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    Quote Originally Posted by Squashed
    This part is a bit confusing. Einstein claimed that a freefalling body was an inertial body (as does the above statement) and so I made the claim that an orbiting satellite was inertial (or near inertial) because it was in continual freefall and yet others claim that it is not inertial because it is in a rotating frame which requires continual change in direction and so it is not inertial.

    publius, do you have any words to clarify my confused state?

    Well, *classically*, that is in terms of Netwonian mechanics, the orbiting body is not inertial. And that's the source of those you were talking with saying it is not inertial.

    For example, standing on the earth, we feel a force, a real force resisting gravity, and I said the proper view (well, the GR view) is really that the ground is accelerating us up. However, you can't have that really happening in 3D space. If the surface of the earth was a balloon that was expanding at an accelerated rate, you'd have that. But then the radius would be expanding and the guy standing on the other side in China would be accelerating away from us, and that obviously ain't happening.

    However, going to 4D (and non-Euclidean 4D) we have more dimensions to work with and, indeed the world line of an observer standing on the surface of the earth is accelerated.

    And, BTW, when thinking about this stuff, forget about the rotation of the earth, the earth/moon orbit, and the orbit of the whle thing around the sun. Just imagine a non-rotating spherical mass in free space, as that other stuff will complicate things greatly. Once you get your mind wrapped around this GR view of that simpler case, you can slowly add in the other effects and see how they play out. For example, the rotation of the earth reduces your weight, the force the ground has to exert to keep you stationary. The centrifugal force subtracts from gravity a little bit. Now, it gets a bit difficult to imagine that in terms of the "real forces". The answer is, rotating with the surface, we're are actually in an orbit, but that orbital speed is too slow to keep from hitting the surface. The ground only has to supply the differential force to maintain the slow circular orbit.

    Oh, well, that's a tangent that makes things complicated, but you've got to keep stuff like that in mind, especially when discussing this.

    Anyway, on the non-rotating sphere, in 4D that real force is accelerating you off the geodesic. But this force does no work. It is similiar to the centripetal force keeping something rotating. It acts only to change direction, not speed, and the kinetic energy is constant.

    And that's what the real force of the ground, pushing up, does in GR. It just continuously changes the direction of your velocity in 4-space, but doesn't change your kinetic energy.

    The component of that trajectory in 3-space, what we can see, is just a straight line.

    An object in orbit is following a geodesic and is really inertial. Because space-time is curved, the projection of that geodesic in 3-space is a closed curve.

    And that path depends on the initial velocity vector, or the angular momentum. If you have enough L, the path is not closed, but a hyperbolic curve. If you don't have enough L, that path hits the surface, and the real force of the ground kicks in quickly and causes a drastic acceleration.

    -Richard

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    I think the others were using "inertial" in the special relativity sense, i.e., a reference frame in which special relativity applies. Gravity ruins special relativity, but you can generalize your definition of inertial.

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    Quote Originally Posted by publius
    Stationary (co accelerating) observers however, would "feel the gradient" across their length. It may be the tidal forces are actually velocity dependent. That is, it would be different at the same stationary point in the gravitational field if I dropped an object from different altitudes.

    This does get complicated.
    I haven't responded to this typically subtle post because I haven't figured out what to say, it's too difficult for me without very serious thought. All I can say is that one may need to decide the appropriate frame to discuss tidal effects-- the frame of the inertial observer, or the frame of the accelerated object? If an inertial observer perceives a constant acceleration everywhere for some extended object, then there can be no length contraction, so the object in its own frame must be being stretched. This suggests to me that tidal effects should be analyzed in the frame of the accelerating object, which means you can't do it in special relativity. And that's where I exit the analysis!

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    Ken,

    Don't feel bad, I would've jumped off that calculation train long before you, I imagine.

    This can get even better, though....

    Imagine first this gravitational case: Someone is standing on roof top and you throw a ball straight up (in vaccum, so we don't have to worry about air drag ), with such velocity so it just comes to stop right next the observer on the roof. At that instant, they are side-by-side, with zero relative velocity.

    Now, switch to the accelerating frame version of that. An observer is accelerating, and some inertial ball is moving at high velocity in the same direction and things are just so that the relative velocity is zero right when they are side by side.

    Now, what about the tidal forces then? Again, the ball is completely inertial, so is feeling no tidal forces whatsoever, and the accelerating observer should (using whatever complex calculations are required) come to that conclusion as well.

    But the ball is right next to him, at zero velocity, and at the instant he should be able to look right at it. There should be no Lorentz contraction, and their clocks should be ticking at the same rate.

    Well wait a minute, that would be for a point particle. The accelerating observer will see a gradient in clock rate across the ball (albeit it very small), but he should see no Lorentz contraction (or is there some gravitational length effects similiar to time dialation at work as well?)

    The inertial ball however should see no gradient in clock rate. This may have to do with simultaneity in a very subtle and complex way, and that may have something to do with tidal conundrum. Maybe in the accelerating "roof frame", all parts of the ball do not stop at exactly the same time, and likewise in the frame of the ball, not all parts of the accelerating frame stop at the same time either.

    Switch that back to the gravitational case, and imagine how that would effect tidal calculations in a similiar case. You're in some frame fixed at some point in a gravitational field, and watching things in orbit go by, and want to know what the actual tidal forces are.

    Different frames will certainly see different forces, but they should all agree on "stretch vs no-stretch", and if an object should be pulled apart in a consistent way.

    -Richard

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    Quote Originally Posted by publius
    Again, the ball is completely inertial, so is feeling no tidal forces whatsoever, and the accelerating observer should (using whatever complex calculations are required) come to that conclusion as well.
    I don't think it can be true that the ball feels no tidal effects (or are you imagining a truly constant gravity, like from an infinite plane of mass?). Remember that objects falling into a black hole feel tidal forces.
    Quote Originally Posted by publius
    Well wait a minute, that would be for a point particle. The accelerating observer will see a gradient in clock rate across the ball (albeit it very small), but he should see no Lorentz contraction (or is there some gravitational length effects similiar to time dialation at work as well?)
    I think there's a length effect too, to keep c constant.

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    Quote Originally Posted by Ken G
    I don't think it can be true that the ball feels no tidal effects (or are you imagining a truly constant gravity, like from an infinite plane of mass?). Remember that objects falling into a black hole feel tidal forces.

    I think there's a length effect too, to keep c constant.
    Ken,

    Ah, I didn't make things clear enough. The ball in the inverse square field would feel tidal forces, no question there. But I'm speaking about the ball in a real acceleration psuedo-frame. That ball would be completely inertial, in no gravitational field at all, and would not feel any tidal forces just because some other observer is accelerating.

    And at instant when they're side-by-side, with zero relative velocity, things get interesting. The only difference between the two is one is feeling a force and the other isn't, and I wonder how the transforms would work to reconcile those two views. The accelerating observer seeing the gradient in clock rate, and the gradient in force, the ball seeing no gradient in clock rate across itself, etc.

    My guess is simultaneity would kick in and make it so not all parts stop at the same time in either frame or something.

    Now, this wouldn't happen the same for a inverse square field. But via the Equivalence Principle, if we could construct some mass distribution so we had a real gravitational field exactly like the accelerating observer psuedo field, there would be no tidal forces there for a free falling object, even though there is a gradient in the g field.

    -Richard

  26. #26
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    Ken,

    This can get really confusing, so let me try to explain my tidal point in a different way.

    Classically we think of tidal forces as going as 1/r^3 for a spherical mass field, and the the total tidal stress across some object comes from integrating the difference in g across the object. For two balls separated by a massless string, the tension in that string would just be the difference in g between the end points.

    And that be seen as the difference between the actual g(r) curve and a constant g line, taken at some point on the object, the center of mass being the most logical.

    But, from all the above, we know a frame accelerating at a constant rate has to see a gradient in the equivalent pseudo g(r) field it sees. In classical mechanics, that g(r) would be constant, a truly uniform field. But because of Lorentz contraction, there has to be a gradient. That gradient will be small for low values of acceleration, but significant for large accelerations.

    But even though there is a gradient in the accelerating frame, we know that inertial objects, free falling from the POV of the accelerating do not feel any such gradient force, as they are completely inertial in flat space-time.

    So some cancellation effect must come into play such that the accelerating observer (and by the Equivalance principle, a stationary observer in a real gravitational field), will calculate there is no real tidal force on the "free falling" objects in that particular g(r) field.

    My hunch is in the graph of g(r) vs the constant g at the center of mass, we should use the actual a(r) of the accelerating observer to get the actual tidal force.

    That is, we would integrate g(r) - a(r) across the object. So the correct tidal force a free-falling object experiences in an inverse square field is this difference function, not just a constant offset.

    And even weirder if g(r) were actually constant, a true uniform field, those forces would actually become compressive, not stretching. That sounds crazy, but it seems to be true because of the gradient of the accelerating frame. If g(r) doesn't increase as fast as the accelerating frame gradient, the free falling observer would experience compressive forces.

    I know that sounds preposterous, but thinking about it, I'm coming to that conclusion. Well, wait a minute, if we had some odd mass distribution, maybe we could make a g(r) that decreased in the direction it was pointing, and that would certainly produce compressive tidal forces.

    -Richard

  27. #27
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    It sounds like you are saying, lets consider tidal effects on a truly inertial extended object from two points of view, that of the object and that of an accelerating observer. The object clearly has no tidal effects, and this must be agreed on by the accelerating observer. So when the accelerating observer sees the object accelerate and length contract, he/she must be able to recognize that the length contraction is not actually a tidal effect. Different pieces of the object appear to get different accelerations, but this is simply a result of looking from an accelerating frame and is not actually a tidal effect (I concur that simultaneity issues must be hiding in their somewhere, masquerading as tidal effects). Is that part of what you are saying?

  28. #28
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    Ken,

    I'm getting so wound up in the "Sully DC" nonsense thread, plus going to get to rambling about DC generators and the EM fields in those in another thread, that I can't think about this tonight. If I forget to get back, just give me a rap on the head.

    -Richard

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