Post subject: Physics questions
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We have the "math challenges" superthread, but questions about physics are not really purely math questions, so I thought about starting a thread about questions about physics. (Either problematic challenges or honest "I'm asking because I don't know" questions are ok.) I have actually several questions, but I thought I'll start with just one: As we all know, the speed of light in vacuum is the same for all inertial frames of reference, and because of that, movement in inertial frames of reference is subject to Lorentz transformations. For this reason if there are two spaceships travelling to opposite directions at speed s (from an external observer's point of view), if ship A measures the speed of ship B as it perceives it, it's not 2*s, but something less (and always less than c). It happens because ship B appears contracted in the direction of the movement. Let's assume that ship A measures the speed of ship B by starting a clock when the bows of both ships coincide, and stops the clock when the bow of ship A coincides with the stern of ship B. For the sake of simplicity, let's assume that, while both ships have the same length, ship A measures the apparent length of ship B to be exactly half of its own length (due to Lorentz contraction), and that the time measurement was exactly one second. This gives some speed s' which is less than 2*s, as mentioned earlier. Now, let's assume that ship A wants to estimate what ship B would measure (by using the same method) for the speed of A in this situation. Since it took 1 second for the bow of ship A to go from the bow of ship B to its stern, and since A measured B to be exactly half the length of A, it would mean by reciprocity that it would take 2 seconds for the bow of ship B to go from bow to stern of ship A. This would seem contradictory (as the ships have the same length and are travelling at the same speed). However, this would be from the time reference of ship A, not the time reference of ship B. If ship A wants to estimate what ship B is measuring, it has to use the time reference of ship B, not its own. If there was a big clock visible on the hull of ship B, which ship A could observe, it would see that it would seem to go at half speed compared to the clock in ship A. Hence by using the clock of ship B, it would take one second for the bow of ship B to go from the bow to the stern of ship A, hence reconciling the speed measurement. However, by using the clock of ship B, it takes only a half second for the bow of ship A to go from the bow to the stern of the ship B. Since this is from the time reference of ship B, it would seem that ship B is measuring a half second for this, rather than two seconds (which is what ship A measured for the bow of ship B to go from the bow to the stern of ship A, when using its own clock). How is this apparent contradiction between two seconds vs. a half second reconciled? (This might be related to the so-called ladder paradox, but I honestly don't know how to apply the explanation to this case.)
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Hmm, I only did 1 year of university physics, so I can't give a full answer, but: One thing to note here is that by trying to estimate what B sees by making measurements from the point of view of A you are assuming that A and B agree on lengths; you also assume that A and B agree on events which happen simultaneously; however, events which happen at the same time but different locations in A's perspective will happen at different times in B's perspective. As an example of what I mean, let's assume that A's hull is 1 metre in its own reference frame (and B is 0.5 m in A's reference frame). A can see that after half a second on B's clock, B has moved 0.5 A-metres. At this point in time, two events happen simultaneously from A's perspective: (1) B's bow is halfway along A's hull, and (2) B's stern is in line with A's bow. However, we know that from B's perspective, B is twice the length of A, and that after half a second of B-time A has moved 0.25 B-metres; B's bow is halfway along A's hull but B's stern is 0.75 metres away from A's bow. Event (1) therefore happens before event (2) from B's perspective, and this occurs because A and B disagree on the relative lengths of the two ships. If they disagree on lengths, then A cannot estimate B's observation of speed by combining A-length with B-time, as the example tried to do. B's observations are based on B-length and B-time. I'm struggling to say exactly where the mistake was in the question, but I think I'm on the right lines. I remember a similar-sounding problem: suppose I have a 2-metre long rod and a 1-metre long slot in a table. I am in the table's reference frame. I accellerate the rod until its apparent length is 1 metre, then I apply a force downwards to push the rod through the hole. Diagram:
        force
       vvvvvvv
    -> ==rod== ->relativistic motion
-table-  slot -table-
Because the rod is the same length as the slot, it will fit through. Now consider things from the rod's inertial frame. The slot is now 0.5 metres long and the rod is 2 metres long. How can the rod fit through the slot? Does the rod go through or not? What precisely do things look like from the rod's inertial frame? [I do remember the answer and will give it; but I'm sure somebody else can give a fuller explanation than I'm capable of.]
Post subject: Re: Physics questions
marzojr
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Warp wrote:
As we all know, the speed of light in vacuum is the same for all inertial frames of reference, and because of that, movement in inertial frames of reference is subject to Lorentz transformations.
[nitpick]This only holds on a local basis due to space-time actually being curved (General Relativity and all), but nevermind.[/nitpick]
Warp wrote:
For this reason if there are two spaceships travelling to opposite directions at speed s (from an external observer's point of view), if ship A measures the speed of ship B as it perceives it, it's not 2*s, but something less (and always less than c). It happens because ship B appears contracted in the direction of the movement.
And because time flows differently for A and B than it did from that external observer relative to which both travel at speed s; if you don't include this, the math comes all wrong.
Warp wrote:
Let's assume that ship A measures the speed of ship B by starting a clock when the bows of both ships coincide, and stops the clock when the bow of ship A coincides with the stern of ship B. [snip]
Ah, a simultaneity paradox. These can be tricky to work out, yes. Here, it is easier with a few pictures: In all pictures, time is on the vertical. The pictures are roughly to the scale of the problem (A views B as having about half its length). There are Minkowski diagrams, by the way. On the center is the situation according to the external reference frame in which ships A and B have the same length and move at the same speed; ship A is red, ship B is blue. On the left is the situation as viewed by ship A plotted on the external reference frame; ship A is orange, ship B is light blue. In gray are the simultaneity lines of A (the more horizontal gray lines) and the lines of constant position of A (the more vertical gray lines). In English, this means 'the points which A considers to be simultaneous' and 'the points which A considers to be stationary'. On the right is the situation as viewed by ship B plotted on the external reference frame; ship A is purple, ship B is green. In gray are the simultaneity lines of B (the more horizontal gray lines) and the lines of constant position of B (the more vertical gray lines). In English, this means 'the points which B considers to be simultaneous' and 'the points which B considers to be stationary'. Note that there is some overlap between A and B in the 3 images in the middle lines; these are where the stern of ship B coincides with the bow of ship A according to ship A (left), the external observer (center) and B (right). In the left and right pictures, you have to explicitly include the Lorentz factor on any distances and times measured along the gray lines (due to Minkowski geometry). As you can see, the situation is quite asymmetrical as viewed from within the ships (not so much for the external observer). If anything is unclear, please say so and I will attempt to explain it better.
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Warp wrote:
Since it took 1 second for the bow of ship A to go from the bow of ship B to its stern, and since A measured B to be exactly half the length of A, it would mean by reciprocity that it would take 2 seconds for the bow of ship B to go from bow to stern of ship A.
No. B is only half the length of A along A's space direction. Along B's space direction, it is twice the length of A.
Warp wrote:
However, by using the clock of ship B, it takes only a half second for the bow of ship A to go from the bow to the stern of the ship B. Since this is from the time reference of ship B, it would seem that ship B is measuring a half second for this
Again, you are using the clock of ship B to measure the time along A's time direction. If you measure the time along B's time direction, you get 2 seconds as expected. Call the proper length of each spaceship l. Then, it is apparent that (in implied SI units): The length of each spaceship in the observer's frame is: Let's define some events: E0: when the bows of both ships coincide E1: when the bow of ship A coincides with the stern of ship B E2: when the bow of ship B coincides with the stern of ship A We will write them as [x t] where x is the distance along the direction of movement, and t the distance in the time direction, from event E0. Ship A will be going forwards along the x axis, and ship B will be going backwards. The transformation from A to observer (and from observer to B) is: From observer to A (and from B to observer): The transformation from A to B: And from B to A: So, the coordinates for each event in each frame is: To calculate the length l' of ship B along A's X axis, one has to solve the equation: Multiply by and we get: It's easy to see that the process is exactly the same for measuring the length of ship A along B's X axis.
Post subject: Re: Physics questions
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marzojr wrote:
If anything is unclear, please say so and I will attempt to explain it better.
I'm having a bit of difficulty in interpreting the diagrams. (I read the wikipedia page about them, which was some new info for me. I learned something new, which is always nice, but I have a bit difficulty with your diagrams.) Let's take the left diagram, in other words, the situation as viewed from ship A. At the lowest distance axis (iow. where the bows of both ships coincide) A's clock is started and thus shows 0. Likewise B's clock (which A can see) is started and hence is also 0 (even from A's point of view, if I have understood correctly). At the second distance axis A's bow coincides with B's stern. As established by the problem, at this moment A's clock shows 1 second. However, I have hard time seeing from the diagram (even with the aid of the right diagram) what B's clock is showing, as seen by A, at this point. (My intuitive deduction, as I stated in the original problem, is that B's clock is now showing 0.5 seconds, from A's perspective, but I don't know how to deduce this from the diagram.) Maybe if I understood this from the diagrams, I could better understand what B's clock is showing at the different stages from A's perspective as well as from B's own perspective, which could help me understand the answer to the conundrum.
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Actually, the left diagram is the point of view of ship A as plotted in the external reference frame. I did this so that times and distances for A and B can be directly compared since they have the same speed in that frame. If it helps, here is the combined diagram: This picture is the combination of the 3 images above, plus 3 more lines. The 3 new lines show, in the 3 different frames, the time that the bow of ship B takes to move from bow to stern of ship A. The yellow line and the pink line show the times as measured at the bow of A and B, respectively, and are on the same scale (as the ships have the same speed in this diagram); the cyan line is in the external reference frame, and is in another scale. If the yellow line is 1 second, then the cyan line is sqrt(3/2) (as HHS stated) and the pink line is 2 seconds (it is roughly twice as long in the image, and doing the math will show this to be the case). If the time interval in question is measured at another point in the ships, you can just slide the respective line keeping the endpoints in the same simultaneity lines (for the cyan line, this means "horizontally"). As you can see, these time intervals measure quite different things. And since they only share one event (the origin), it doesn't really make sense to compare them. Now we go to the clocks. If we are going to assume that the folks at ship A are smart enough to check the clock of B at both relevant events and measure the delta, this question can be answered objectively. Otherwise, it will depend on where the clock is located in ship B. Assuming smart folks, we can place the clock somewhere convenient -- say, at the stern of ship B. Then, we have a Minkowski triangle with two known sides (the yellow line and the lowest light blue line) and one unknown side (the gray line between them). This unknown side is the time interval elapsed in the bow of ship B between the two relevant events as seen by A. Now, the side can be calculated if we have the rest length of the ships. I will represent this length by L (this is the length of the orange and purple lines). We know that whatever this rest length is, the light blue lines (and the green lines too) will be half of this: L/2. The yellow line has "length" equal to c * 1 s. Putting in the proper time formula (which is a special case of the space-time version of Pythagoras theorem), we have that the desired time interval dt is dt = sqrt[(1 s)^2 - (L/2c)^2] This is how much time will elapse in a clock on ship B as seen by ship A in the problem at hand. I hope this is clear enough; this kind of thing is much easier to explain in person, having instantaneous feedback from the subject if something is unclear...
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I have to confess that I have hard time getting my head to understand the diagram and how the locations and times work from the perspective of the different observers... :/ Thinking about it independently, the answer might lie on defining more precisely some of the things which were imprecisely stated in the original problem description. As you mentioned, the precise location of the visible clock on ship B may, in fact, have a radical influence on the outcome of the measurements ship A is making. I didn't precisely specify where the clock is located because I didn't think about it. The logical place for the visible clock on ship B is the bow of the ship. This way when the bows of both ships coincide, everything should be completely "synchronized": A sees that its bow coincides with B and starts its clock, and B sees that its bow coincides with A and starts its clock, and the visible clock on B should at this moment read 0 from both A's and B's perspective (if I have understood correctly, because if we think of the problem as a one-dimensional one, both bows and the visible clock are at the same place at the same time). Another probably important detail which I inadvertently left out in the original problem is where exactly A is making the measurements from. The logical place for A to make the measurements is at the bow of ship A (because that's the place where it can see when both bows coincide, as well as when A's bow coincides with B's stern). So now when A sees that its bow coincides with B's stern, and looks at B's clock to see what it reads, this clock is actually a half-ship away from where A is making the measurement from (ie. A's bow). Here's probably where the relativity of simultaneity kicks in: A is seeing these two events as simultaneous: 1) A's bow coinciding with B's stern 2) B's bow being at half of A's length, and B's clock reading something (0.5 seconds?) However, from B's point of view these two events are not simultaneous, so even if its clock does read 0.5 (or whatever it should read at this point), its stern is not coincident with A's bow (from B's point of view). This is because the place where A is making the measurements is not coincident with where B is making the measurements, at this point in time (and because they are in different frames of reference). One could perhaps say that A is getting a "false reading" from B's clock (because A is not compensating for the relativity of simultaneity that is happening, but taking the reading as-is). Your diagram is probably telling this same story, but as said, I have hard time understanding it... :/ Edit: This raises another interesting question: If ship A had a measurement device at the half point of the ship which looks what the clock at B's bow is showing (at the moment when B's bow is exactly at this half point of ship A), would it see a different value than the measurement device located at A's bow? If yes, what would the two readings be?
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Warp wrote:
If ship A had a measurement device at the half point of the ship which looks what the clock at B's bow is showing (at the moment when B's bow is exactly at this half point of ship A), would it see a different value than the measurement device located at A's bow? If yes, what would the two readings be?
If there is a clock at B's stern synchronized in B's time with the clock at B's bow, it will show 2 seconds when the clock at B's bow is showing 0.5 seconds. It will always be 1.5 seconds ahead of the clock at B's bow as seen from A.
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@Warp: A more in-depth explanation: Minkowski diagrams are very similar to standard x-y graphs: you read off points as you would read in any such plots -- to get the t coordinate in any given reference frame, you trace a line parallel to that frame's simultaneity line until it reaches the time axis for that reference frame and read the value at the time axis (and vice versa to get the position). The only difference is the distance function -- in a x-y graph, you use Pythagoras theorem to get the distance along slanted curves, in a Minkowski diagram, you measure the space-time interval given by the Minkowski metric. I would link to the Wikipedia page on this, but it is highly technical; so the short version: In space-time, you can define a "distance" between any two events by the following formula:
S = (c * delta-t)^2 - (delta-x)^2 - (delta-y)^2 - (delta-z)^2
If S > 0, you have what is called a timelike interval, and sqrt(S)/c is the proper time interval that elapses on an inertial reference frame that considers these two events to have the same spatial coordinates -- and there will be such frames (this would be an absolute "temporal distance" between the two events). Moreover, you will have a fixed, definite, temporal ordering of these two events -- there will not be any reference frame (inertial or otherwise) which considers these two events to be simultaneous or which reverses the order in which these two events happen (this would require either traveling back in time or superluminal speeds). Conversely, you cannot define any absolute spatial distance between these events. If S < 0, you have a spacelike interval, and sqrt(-S) is the proper distance between the two events in any inertial reference frame that considers them to have the same time coordinates -- i.e., simultaneous (this will be an absolute spacial distance between the two events). You cannot define any absolute temporal order between these two events, and there are reference frames (inertial or otherwise) in which either event may happen before the other; moreover, you cannot find any reference frame (inertial or otherwise) in which these two events are at the same spatial location (this would require superluminal speeds). If S = 0, you have a null or lightlike interval -- the interval is zero, and this means that the two events can be connected by a beam of light. The Pythagorean distance is invariant with regard to rotations of the coordinate systems (as well as translations); likewise, the Minkowski interval is invariant with regards to purely spatial rotations, as well as rotations in a plane involving the time axis -- these are called Lorentz transformations. If you add translations of the coordinate system to the Lorentz transformations, you reach the Poincaré transformations, which also preserve the Minkowski interval. All of this is based on the prevailing modern view of Relativity as being a study of the geometry of space-time, by the way; and while, historically, the Minkowski interval was derived from the Lorentz transformations, it is now known today that it is the interval -- the space-time metric -- which is the important thing. In particular, according to this distance function, you have one important consequence (which I am not going to prove): return to the first 3 diagrams I posted and look at either the left or right diagrams. In these diagrams, the more vertical gray lines (the ones which have arrows in the last diagram I posted) are orthogonal to the more horizontal gray lines. That is a Lorentz transformation, or a rotation in space-time for you :-) ----------------------- So, with all of that in mind, let us return to the last diagram I posted. If you notice, some of the gray lines have arrows in them, as does one of the black lines. These are the temporal axis of the moving ships. While I drew two for each ship, one for the bow and one for the stern of each ship, you only need one -- any one, including any other you wish to draw which is parallel to these. Selecting one means selecting what position each ship considers to be position x = 0 in its frame. Lets pick the time axis on the bow of each ship. The gray lines without arrows, the simultaneity lines, work similarly. You can pick one to indicate what you consider the time t = 0 according to that frame (including one parallel to the ones I drew). Lets pick the lower gray lines to be the time t = 0. Combined with the above, all 3 reference frames have the same event as the origin of their coordinate systems. Hopefully, now things are starting to become more clear. So I will now start some analysis of the diagram to illustrate how it goes. I will assume that the ships have proper lengths equal to L from now on. First, the point where the two black lines meet -- the origin. This point has the coordinates (t, x) = (0, 0) in the 3 coordinate systems. Now, the yellow line has two endpoints: the origin and the point where the stern of B coincides with the bow of A. Now, the bow of A is always at position x = 0 according to A, and in the problem, we assumed that this happens when the clock at A marks 1 s. Thus, the other endpoint is at (t, x) = (1 s, 0) according to ship A. The Minkowski interval tells us that this is a proper time interval, and the time that elapses in A is 1 s -- which we already knew. Moving this to other reference frames will be tricky yet, so I will leave it for later. Take now the orange lines: they have as one endpoint the bow of ship A and ad the other endpoint the stern of ship A. At this point, we can obtain the coordinates of the endpoints of all of the orange lines in the reference frame of A: the lowest orange line goes from (0, -L) to (0, 0), while the middle orange line goes from (1 s, -L) to (1 s, 0). Both of these come from the description of the problem. Now to find the endpoints of the topmost orange line: the defining characteristic is that the stern of ship B has moved enough (from the point of view of ship A) that it now coincides with the stern of ship A. By hypothesis, ship B has then moved (according to ship A) 3 times its length (in A's point of view) -- since it took B 1 s to move its own length (again, according to A), then the topmost orange line has coordinates going from (3 s, -L) to (3 s, 0) according to A. And so on. The trickiest thing to calculate are the speeds of A and B according to the external reference frame, as well as the gamma factor for this speed; this is because you need to calculate the hyperbolic angle and use inverse hyperbolic tangents and so on (see this for something about hyperbolic functions). -------------------------- Now, there is what HHS said. You can (hopefully) verify it from the diagram now.
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I understood your explanation, and it makes it clear what A is measuring using its own clock. However, I still have hard time deducing the answer to the original problem: Let's assume there's a visible clock on the bow of ship B, which A can see during the entire experiment. What does someone located at the bow of ship A (or at some other point in ship A) see as the value of this clock at different times, and how does this relate to what someone on ship B sees (when looking at his own clock)?
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marzojr wrote:
And so on. The trickiest thing to calculate are the speeds of A and B according to the external reference frame, as well as the gamma factor for this speed; this is because you need to calculate the hyperbolic angle and use inverse hyperbolic tangents and so on
Actually, no calculation of angles is necessary, just a plain old quadratic equation: One can easily see, by writing the last two equations in terms of λ' and factorizing, that in general in this situation with an observer positioned in the middle between two ships according to his own inert frame, one can calculate the relative speed and λ between each ship and the observer from the λ' between the two ships in the following way:
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In order to solve that, you need to compute what time has elapsed in B's clocks; the time intervals will, of course, be different based on the position of the clock, because A will "see" things in his own simultaneity lines. Which brings me to a small quibble -- in Physics, particularly Relativity, it is customary to distinguish between "seeing" and "observing". "Seeing" something means that the light from the object has reached your eyes; "observing" something means that, after seeing it, you subtract the light travel time and distance in order to work out the event from which the light came from. Usually, what you see is different from what you observe. So I will assume that you are asking what (the bow of) A will observe in B's clock; this will be easier to work out. Now, back to the problem: to work out what A observes each clock of B to be at any given instance of A's time, we have to work out how much time has elapsed in B since its t = 0 line and until the desired events which A considers to be simultaneous. First, lets work out how much time has elapsed in B's stern from B's time t = 0 until A's time 1 s (when the bow of A coincides with the stern of B). This is easy, because we already know it: it is 2 s. This is the length of the gray line that goes from the bottom-right end of the lowest purple line until the rightmost point of the yellow line. A quick aside: I have been using L for the (rest) length of the ships. I just realized, however, that the problem allows us to obtain how much it is. We know that, in ship A's point of view, ship B took 1 s to pass the bow of A, and that the length of B in A's frame is half the length of A -- the gamma factor between them is 2. The first info means that the speed of ship B in A's frame is L/2s; plugging this into the formula for the gamma factor gives
2 = gamma = 1/sqrt(1 - (L/(2s*c))^2)
4 - (L/(1s*c))^2 = 1
L = sqrt(3) * 1 s * c
or sqrt(3) light-seconds (about 5.2E8 m) -- those ships are HUGE! (there is a way to obtain this from the diagram as well, but I will leave it as an exercise :-p). Now to the bow of B from B's time t = 0 until A's time of 1 s. This will be the length of the stretch of pink line going from lower-left point of light blue line until the lower-left point of middle light blue line. We know the coordinates of these points in A's frame -- one is (0, 0), while the other is (1 s, -sqrt(3)/2 light-seconds). So the time interval between these two is
dt = sqrt((1 s)^2 - (-sqrt(3)/2 s)^2) = sqrt(1 - 3/4) s = 0.5 s
So at time t = 1 s according to ship A, the clocks at the bow and stern of ship B were observed to be, respectively, 0.5 s and 2 s. This 1.5 s difference will be maintained by the clocks of B as observed in A's frame because they travel with the same speed.
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HHS wrote:
Actually, no calculation of angles is necessary, just a plain old quadratic equation:
You are, of course, correct; in my defense, I was too tired when I wrote that :-)
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Warp wrote:
I understood your explanation, and it makes it clear what A is measuring using its own clock. However, I still have hard time deducing the answer to the original problem: Let's assume there's a visible clock on the bow of ship B, which A can see during the entire experiment. What does someone located at the bow of ship A (or at some other point in ship A) see as the value of this clock at different times, and how does this relate to what someone on ship B sees (when looking at his own clock)?
At time t according to A, B's bow clock shows t/2 and B's stern clock shows t/2+1.5. The time difference is equal to the relative speed times the distance (as measured by B) divided by c squared. What A actually sees depends on the distance between A and B, since the light has to travel from B to A before A can see it. Using the following variables: u: Observer's time coordinate according to A v: Observer's space coordinate according to A t': Observed clock time t: Corresponding A's time x: Relative distance in the direction of movement between A and B y: Relative distance in the perpendicular direction between observer and B s': Relative speed λ': Relative gamma Then we'll have: This can also be written as: If we plug in our gamma, we get: This gives B's clock time as seen by an observer at [v u] in A's frame. As for B, he will of course at his coordinates [v u] simply see his bow clock time as .
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Suppose you have a ramp at an angle alpha from horizontal, and a cube of mass m and density d on the ramp. You can assume that this is all in vacuum. When let go, the cube slides a distance l in t seconds. Now, assume that instead of a cube it's a sphere with the same mass and made of the same material (iow. it's of the same density). When let go, will it take a shorter or a longer time to move the same distance, or the same time? Why? (And if you are up to it, how much faster/slower?)
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There are way to many possible results depending on the friction between the ball and the surface. If there is no friction, the ball will not roll and will travel at the same speed as the cube did. If there is friction, the ball will roll, and part of the kinetic energy will be rotational -- thus slowing the ball. It then depends on whether or not the friction is high enough to prevent sliding or not. If the ball does not slide as it rolls down, the ball will be at its slowest. And it is the easiest to compute. Lets call the side of the cube a = (m/d)^(1/2) and the radius of the ball r = (3*m/4*pi*d)^(1/3). Now, since the ball did not slide, then the speed of the point of contact of the ball and ramp must be purely radial (a tangential component would be sliding). With all that, we can set up our coordinate system. The best choice is one in which the ball moves along one axis (say, towards the positive the x direction) parallel to the ramp. The slanted "vertical" (y) direction will have no motion, and we can ignore it. The other important variable is the radial angle q that describes the rolling motion of the ball. The no-sliding condition can be written as r*dq/dt + dx/dt = 0 The motion on the x direction comes from Newton's equation: d^2x/dt^2 = g*sin(alpha) - Fn/m And so does the rolling: I*d^2q/dt^2 = r*Fn Here, I = (2/5)*m*r^3 is the moment of inertial of the ball around its center and Fn is the frictional force, which acts as a torque. Putting it all together to eliminate Fn, we have (I/m*r)*d^2q/dt^2 = -d^2x/dt^2 + g*sin(alpha) d/dt {(I/m*r)*dq/dt + dx/dt)} = g*sin(alpha) Using the no-sliding condition, ((I-m*r)/m*r)*d^2q/dt^2 = g*sin(alpha) Integrating once in time, we have dq/dt = (m*r/(I-m*r))*g*sin(alpha)*t + c1 where c1 is a constants of integration; if the ball starts with zero angular speed, c1 = 0. Returning to the no-sliding condition, we have dx/dt = m*r^2*g*sin(alpha)*t/(I-m*r) x = (1/2)*m*r^2*g*sin(alpha)*t^2/(I-m*r) + c2 where c2 is a constants of integration; if the ball starts at x = 0, c2 = 0. We know that the ball has traveled a distance L along the x axis; hence, we can solve for t: t = sqrt{ 2*L*(I-m*r)/m*r^2*g*sin(alpha) } I will leave substituting I and r back, as well as comparing with the cube, for you.
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What about the scenario where the cube also rolls?
marzojr
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That one is a lot more complicated. I did the math once for the initial part of a similar system when I was teaching undergraduates in a lab -- and it is ugly. The system in question was regarding an experiment with an inclined plane of variable inclination and stacks of cylindrical weights (they had a hole in the middle and a stick to prevent the top ones from sliding from the other cylinders). The students had to analyze the sliding angle of the cylindrical weights when changing the number of stacked cylindrical weights, and the tilting effect caused the higher stacks to slide at lower angles than would have been expected otherwise, as the tilt reduced effective friction. I don't know where my notes are (it was a few years back), but it was done with the method of Lagrange multipliers in classical variational calculus. But the gist is that the cube will tilt first due to the inclination and friction; this will reduce the normal force, which reduces the friction. This will cause the cube to slide. If the friction is still high enough (or the plane inclined enough), the cube will tilt faster than it slides. From then on, the cube will bounce on the inclined plane when it finishes the first 90-degree rotation, so it depends on the elastic properties of the cube and of the inclined plane. I didn't go that far in the math, though, as I had already seen what I wanted -- the reduction of the sliding angle for the cylinders.
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The General Theory of Relativity predicts that if a particle enters the event horizon of a black hole, it would require an infinite amount of energy to stop it from falling into the singularity at the center. On the other hand, the Pauli exclusion principle states, basically, that you would need an infinite amount of energy to make two particles be at the same place at the same time (which is what is happening in a singularity). How are these two opposite forces consolidated? (And please, I would prefer answers based on science instead of answers based on personal incredulity.)
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Warp wrote:
The General Theory of Relativity predicts that if a particle enters the event horizon of a black hole, it would require an infinite amount of energy to stop it from falling into the singularity at the center. On the other hand, the Pauli exclusion principle states, basically, that you would need an infinite amount of energy to make two particles be at the same place at the same time (which is what is happening in a singularity). How are these two opposite forces consolidated?
They aren't. Not with current theories. General relativity and quantum mechanics (of which Pauli exclusion is a part) simply do not agree, which means than both are incomplete descriptions of the universe. The sheer inability to observe the interior of a black hole makes it difficult to measure what actually happens; perhaps if we knew, we'd be closer to solving this great unsolved problem of physics. What this means in practice is that, according to quantum mechanics, gravity does not exist. It's not in the models at all. There are only three forces: electromagnetism, weak and strong. According to general relativity, all particles are classical. There's no wavefunctions, no wave-particle duality, no quantum effects. Obviously, these are slightly embarassing omissions from the respective theories; but if you operate on a scale where the omissions are negligible (ie the very small for QM, the very large for GR) it doesn't matter too much. The field of physics trying to unify these theories is called quantum gravity. You'll be waiting a while for an answer to your question. (The wp article lists "Points of tension" under which it says "however, no one is certain that classical general relativity applies near singularities in the first place"; reflecting the general problem of describing something which no information can escape from.)
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Warp wrote:
The General Theory of Relativity predicts that if a particle enters the event horizon of a black hole, it would require an infinite amount of energy to stop it from falling into the singularity at the center.
Not quite; ignoring for a moment the distinct types of black holes, and assuming a Schwarzschild black hole (non-rotating, uncharged), then no amount of energy will allow you to scape once you get past the event horizon. This is because any amount of energy pumped into acceleration (even "infinite" energy) will still leave you moving inside the local light cone. Infinite energy is also a terrible thing in GR -- energy curves space-time too, and having an infinite amount of it means you are a black hole... Moreover, contrary to normal intuition, the harder you try to accelerate away from the singularity, the faster you are going to hit it according to your proper time. Other than that, rhebus is on the money: the biggest problem in Physics today is that the best theories it has for explaining the Universe are mutually incompatible at a fundamental level.
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rhebus and marzojr gave great explanations. I'll make a minor addition by pointing out that it's a bit naive to say things like "you would need an infinite amount of energy to make two particles be at the same place at the same time (which is what is happening in a singularity.)" First of all, no scientist, when being precise, would ever talk about "an infinite amount of energy" or some such. For example, you've probably heard laypeople say things like "if you had an infinite amount of energy you could accelerate a spaceship to the speed of light." That's not a meaningful statement because there is no such thing as an infinite amount of energy, and it certainly doesn't follow from relativity. What relativity tells us is that no finite amount of energy is sufficient to accelerate a massive object to the speed of light. That certainly doesn't imply there is such a thing as "infinite energy." Second, it's dangerous to conflate to such a degree "the way things are" with "the way we describe things." As rhebus pointed out, GTR and QM are two ways we have of describing the universe, though they are both neither complete nor even comparable in many meaningful senses. We use them to describe different portions of the universe, and they aren't often useful (or even meaningful) outside of their limited domains. When physicists talk of singularities and such, it doesn't actually mean they believe there are precise physical manifestations corresponding to the ways mathematical models fail. That is, a "singularity" hidden by the event horizon of a black hole isn't a "thing" as far as anyone can know. It's just a situation where scientists admit our best methods of describing the universe are not sufficient. So to say "[something] is what is happening in a singularity" is wrong simply because it's meaningless. Nothing is known to happen in a singularity because it's not a thing, and our current methods of describing the universe are not sufficient to describe situations that, when they arise, result in mathematical singularities in our models.
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Pointless Boy wrote:
First of all, no scientist, when being precise, would ever talk about "an infinite amount of energy" or some such. For example, you've probably heard laypeople say things like "if you had an infinite amount of energy you could accelerate a spaceship to the speed of light." That's not a meaningful statement because there is no such thing as an infinite amount of energy, and it certainly doesn't follow from relativity. What relativity tells us is that no finite amount of energy is sufficient to accelerate a massive object to the speed of light. That certainly doesn't imply there is such a thing as "infinite energy."
To a layman (like me) there is little difference between the concepts of "you would need an infinite amount of energy to" and "no finite amount of energy is enough to", as they sound like two ways of expressing the same thing. As you say, though, there's probably a difference, at least technically speaking. (It's probably similar to the difference between "an infinite amount of RAM" and "unbounded/unlimited memory" in computer science. They might sound like the same thing to a layman, but there's a technical difference.)
When physicists talk of singularities and such, it doesn't actually mean they believe there are precise physical manifestations corresponding to the ways mathematical models fail.
When I wrote "which is what is happening in a singularity" I was implying "according to GR". (In other words, pure GR predicts that inside a black hole it's impossible for a particle to keep out of the singularity and hence the only possibility is for all matter to compress into a zero-sized point. Of course GR doesn't take into account that according to QM it's impossible for a singularity to happen, which is what causes the dilemma.)
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Warp wrote:
To a layman (like me) there is little difference between the concepts of "you would need an infinite amount of energy to" and "no finite amount of energy is enough to", as they sound like two ways of expressing the same thing.
These are not logically the same. One place where these necessarily differ is where even an infinite amount of energy won't do what you want to achieve. For example, if you take the special relativity equations and dumbly plug infinite energy in (more accurately, if you take the limit as energy goes to infinity, since otherwise you can't deal with infinities in equations in a well-defined way), you find that a massive particle with infinite energy travels at the speed of light. No finite or infinite amount of energy is enough to travel faster than light. IOW, just because a finite amount of energy couldn't do the job, doesn't mean that an infinite amount of energy will. Even if you take the limit of infinite energy (and ignore the curving of spacetime that marzojr mentioned this would cause) a massive particle can't escape a black hole -- since a massive particle with "infinite energy" moves like a massless particle, and massless particles can't escape a black hole either. Your statement "it would require an infinite amount of energy to stop it from falling into the singularity at the center" isn't true -- even an infinite amount of energy wouldn't be able to stop it.