Hello all. First time posting here. I have a basic understanding of special relativity and I have few questions about what the Large Hadron Collider (LHC) looks like to a proton going around inside it. As I thought more and more about it the few questions turned into a bunch of questions. I know that special relativity might not apply everywhere since the proton is accelerated by traveling in a circle. I am not sure where special relativity can be used and where general relativity would be used. What shape does the LHC have to a proton traveling the 27km circumference with a Lorentz factor of 7500 (in about 89 ns)? I am guessing it would look like a very flat oval with a diameter contracted by 7500 in the direction the proton is moving and the diameter perpendicular to that (about 8.6km) would not be contracted. Is that correct? At least, that is what it would look like to a proton traveling straight and tangent to the LHC with the same speed. Does traveling in an arc make a difference? By the way, assuming it's an oval, is the oval an ellipse or some other shape? (When I say "looks like" I mean according only to whatever relativistic transformations would apply, not what it looks like to a camera, but that would be another interesting question.) Also, would a proton see clocks on the circumference of the LHC running slower or faster? The moment a proton passes by a clock on the circumference it seems that special relativity would say it would be running slower. Is that correct? But it also seems that when a proton makes its 12ns (89us/7500 I'm assuming) cycle according to the proton's clock, an LHC clock on the circumference would have to show 89us having passed when a proton passes by a second time making the clock appear to run faster. How is this resolved? How would a clock at the center of the LHC or on the opposite side of the LHC appear to run to a proton? What about a proton's clock on the opposite side? How long is the 27km LHC circumference as seen by the proton? It seems it could be 12 light nanoseconds (3.6m) considering the proton sees itself traveling in the LHC at near the speed of light for 12ns to make a cycle, or it could be about 2 x 8.6km = 17.2km looking at the very flat oval with the circumference being about two non-contracted diameters, or it could be 27km since it seems the proton is always at the most outer point of the oval from its point of view. I'm not sure if this is a meaningful question but, if a proton were to slowly feed out a string, with 1 meter markings, behind itself as it went around, how much string would be fed out, by the markings, when the string gets all the way around? I am guessing that it could be 27km x 7500 = 20,2500km, calculating from the LHC point of view if the string is contracted. If the proton looks across the LHC to the other side, what marking does it see on the string. It seems it would be the halfway mark, or could it be something else? Are the 1 meter string markings on the opposite side contracted even more than 7500 since the string is traveling in the "oncoming" direction? If so, how much? From the four different numbers the circumference, are some more correct or meaningful than the others? Also, what radius of curvature and centrifugal force (how many g's) does the proton experiences as it goes around? And finally, what would everything look like to a proton's video camera?
Hi, welcome to PF :) Yes, the shape is an ellipse and at the tangential velocity specified the width of the ellipse would be just over one meter. Correct, the proton see the clock on the circumference as running at a 7500 times slower rate as it passes. An observer on perimeter would say a clock moving with the proton is running 7500 times slower rate than a clock on the perimeter, as the proton clock passes close by. Each time the proton completes a circuit it will find that its elapsed time is 7500 times less than the elapsed time of the clock on the perimeter even though the instantaneous clock rate of the perimeter clock seems to 7500 times slower. The difference is between instantaneous clock rates and accumulated elapsed times. It can also be noted that the circulating proton feels very strong centripetal forces which makes it feel like it is in a very strong gravitational field while the clock on the perimeter does not. As mentioned above the proton feels very strong centripetal forces. Actual experiments with cyclotrons have shown that the time dilation of muon half lifes is exactly what you would expect if you only consider the instantaneous tangential velocity. On the other hand, from the point of view of a muon inside the cyclotron, the muon is stationary and it feels like it is in a very strong gravitational field and it could reason that its own clock is running slow due to gravitational time dilation relative to say a clock very near the centre of the cyclotron. In fact, it would see the clock near the centre as stationary but light coming from the central clock would be blue shifted from the point of view of the circulating clock and this could appear to be a gravitational effect. One important point is that time dilation can be put down to velocity as per Special Relativity or due to acceleration, but care has to be taken not to double count. It is a meaningful question and actually it is a good question too ;) Yes, the string would be 202,500km long according to the proton and 27km long according to observers at rest with the LHC. If the proton was to suddenly stop the LHC observers would also measure the length of the string to be 202,500km long when the string comes to rest in their frame. Now if the circulating proton considers itself at rest, and measures how long it takes for a mark on the perimeter of the LHC ring to complete a circuit it will calculate the length of the LHC ring to be about 3.6 metrs and it will put this down to length contraction of the ring due to the motion of the ring relative to the proton. It could also calculate that the proper length of the ring is 27kms when length contraction is allowed for. It would be slightly less than the halfway mark because light arriving at the proton from the other side of the LHC would have left at an earlier time because of the finite speed of light. Yes they are and you will have to look up "relativistic velocity addition" to work that out. They all are valid in their own ways, but most people would argue that the most meaningful measure of length is the "proper length" which is the length measured by an observer at rest with object being measured. In this case 27km is the proper length of the LHC ring and 202,500km is the proper length of the string. Just because the moving string wraps neatly aroung the LHC circumference does not mean that the string is a meaningful measure of the LHC perimeter length. Different observers circulating around the LHC at different velocities with their own string measures would all obtain different measures of the length of the perimeter. However, if they calculate the length by how long a mark takes to goes round multiplied by the velocity of the mark and then allow for for length contraction they would all agree on the proper length of the LHC perimeter.
Teve, General Relatvity is irrelevant to this question. Special relativity just does the job from A to Z. It is a widespread misconception that SR cannot deal with acceleration. Actually, this is even not a misconception, this is plain wrong. Michel
If I did my math right, I think the instantaneous length contraction and time dilation for two protons at near light speed, each with Lorantz factor 7500 is approximately 15000. I think that would apply for instantaneous clock rates for nearby oncoming protons. Is that correct? I think it would also apply to the marked meters on the string. But two oncoming protons should encounter each other twice a cycle and both should see their clocks showing 6ns having elapse since their last meeting. It seems that observed clock rates have to run faster for clocks while they are separated before they later meet up again. Also, does a counter circulating proton looking across the LHC see non-contracted meter markings on the string since they are moving in the same direction? Back to clocks. What clock rate does a proton see (not by camera but by relativistic transformation) for protons on the other side going in either direction, and what clock rate does the proton see for an LHC clock on the other side and at the center of the LHC? Counter circulating protons on opposite sides would have the same instantaneous velocity, but then they are separated in their "gavitational" fields. For the center LHC clock it seems the clock rate the proton sees is 7500 times faster than its own. How much elapsed time is seen by a proton on a proton clock or LHC clock when it is on the other side? (Again, not by camera but by transformation.) It seems that counter circulating protons would each have to see 3ns (1/4 cycle) of elapsed time on each others (and their own) clock when they are opposite each other just after they last met. If two counter circulating protons synchronize their clocks to 0 when they meet, as a function of the proton's time, what is the time on the other proton's clock? I don't think they read the same (from one proton's view) shortly after the meet, but they will be synchronized when they meet again and when they are opposite each other it seems. It seems it has involve a periodic function of some sort. Protons circulating in the same direction and on opposite sides are "oncoming" but never meet. It seems they can synchronize using the center LHC clock. How do they see each others clock rates? It seems they would both see the same increased clock rate for the center clock. This seems to suggest that proton clocks always on opposite sides, and all proton's circulating in the same direction for that matter, see each others clocks running at the same rate. Is this right? But is seems the same can be applied to the counter rotating protons so that all proton clocks run at the same rate relative to each other. This contradicts the idea that oncoming protons near each other see different clock rates between them. What am I missing? I have heard that special relativity can be used for "no gravity/mass" situations but I am not sure how it would be applied to all the questions above. Specifically the LHC circumference clocks and maybe the counter circulating clocks which are time dilated when near by but have to be the reverse of time dilated (expanded?) at other times. Also, one more question. If the LHC were on a very dense planet with protons accelerated to a Lorentz factor of 7500, and I were hovering above it far away so that the gravitational time dilation was, say, a factor of 10, is the overall time dilation I see for the protons just the product of the two, or 75000, or is the general formula more complicated?
Hi kev, you need to be a little careful here. The proton's rest frame is a non-inertial frame, as you point out later. Both the proton and the perimeter clock agree that over the course of one "lap" the proton's clock is the slow one. Once you try to compare the clock at other points besides once per lap when they are right next to each other you get into all sorts of hairy simultaneity issues. I am sure that you could indeed show what you are saying here, but it would be complicated.
Why should it be complicated? It's standard time dilatation in the momentary comoving inertial frames.
Even just that is already complicated because the definition of simultaneity is continuously changing.
Hmm.. I'd say that kev is talking about a snapshot, where the proton's motion can be approximated by a straight, inertial line: I think kev was more than a little carful, his wording is quite explicit. (unintended pun)
For a Lorentz factor of 7500 the velocity is aprox (1-9E-9)c while the combined velocity of two oncoming protons is aprox (1-4E-17)c using the relativistic velocity addition formula. The time dilation factor of the combined velocity is about 112,499,999. Spreadsheet software can not handle this sort of precision. You need something like Pari GP. Each would see the marked meters on the other oncoming proton's string as about 112 million times shorter than their on their own string and yes each would see that an equal amount of time of about 6ns of time has elapsed on both their clocks when they next meet despite the fact they both perceived each others clocks to be running 112 million times slower than their own clocks when they pass by each other. Lets make it slightly simpler by making the LHC more like a race track with two long parallel straights close to each other and tight bends connecting the ends. Counter rotating protons passing close to each other on the long straights (but on opposite side of the track) would see no length contraction in the string of the other proton on the other side of the track because as you say, at that point they are co-moving. If the string of the proton on the other side is long enough to extend all the way around then the part of the string that is on the side of the track that is opposite to the proton that it is attached to will be length contracted by a factor of 112 million according to a counter rotating proton that is alongside that part of the string. As hinted above, a proton on one straight would see the clock rate of a counter rotating proton on the opposite straight, ticking at the same rate as its own clock (as it passed close by). It would see the clock rate of co-rotating proton clock on the opposite straight as ticking 112 million times slower than its own clock (as it passed close by). It would also see the clock rate of a clock at the centre of the track as ticking 7500 times slower than its own clock as it passes close by it on a long straight. On the other hand, when the proton is furthest from the centre clock at one the curved sections it would perceive the centre clcok to ticking faster than its own clock, but at that point the proton would be aware of strong centripetal forces. In the circular track scenario, protons on opposite sides are in the same “gravitational field” because they feel the same acceleration and are at the same “gravitational radius” from the centre. OK, circular track again. You do not want purely visual effects but we should at least consider what they may be, so they can be discounted. First there are effects due to light travel times so that when the proton sees an object on the opposite side of the track, that object has moved on so it is seeing an image from an earlier time makes the apparent elapsed time on the opposite clocks appear less than you might otherwise expect. Now we come to Doppler effects. For two protons circulating in the same direction but on opposite sides, the calculated transverse Doppler shift is a redshift factor of 112 million when we consider the instantaneous transverse velocities. However, transverse Doppler shift normally considers two inertially moving observers where each observer is time dilated in the frame of the other observer. In this situation it can be seen that in the LHC frame the clocks of the two protons are ticking at exactly the same rate as each other, so if one proton transmits signals at a rate of one per second the proton on the opposite side should receive those signals at a rate of one per second when the frequencies are measured using one receiving clock. In other words my guess is that protons circulating in the same direction will not see a Doppler shift in the clocks of the protons on the opposite side. If they calculate frequencies by using two nearby co-moving and synchronised clocks then would observe mutual time dilation. The reason this apparent contradiction arises, is because when you try to synchronise clocks moving in the same direction in a circle it turns out to be impossible to synchronise all the clocks with each other all the way around, which is a consequence of the fact that a non inertial reference frame can not be treated as an inertial frame on a large scale. In the LHC frame the centre clock is ticking faster than the circulating proton clocks and in my opinion the circulating protons will also see the centre clock as advancing at a faster rate and this could be interpreted as being due the centre clock being higher in the effective “gravitational field”. Another visual effect to take into account is aberration and to a circulating proton the centre clock (or a proton on the opposite side) will not be appear to be at right angles to its instantaneous direction of travel, but would appear to be somewhere towards the front of its field of view when looking in the direction it is travelling. Now as for what they consider the elapsed time to be, of a co-circulating clock on the opposite side, it is hard to be definitive. This is because as I mentioned before, if they synchronise a series of nearby and approximately co-moving clocks they will obtain different results depending on which direction they synchronise the clocks in. They could of course note that they measure the same elapsed time for each circuit they complete and simply assume that their mutual elapsed times must be the advancing at the same rate. Knowing the “real elapsed time” of a distant clock is not always a trivial matter and this question probably requires a lot more thought than I gave it my answer above. If we revert to the race track model with parallel straights, then if two counter rotating protons synchronise clocks when they pass each other on a bend, then when they are alongside each other on a long straight they will measure each others clocks to be advancing at the same rate. They will also show the same elapsed times but when they cross each other at the far bend they will each measure the clock rate of the other proton to be slower than their own clock as they pass close by each other, (but they will still be showing the same elapsed times as each other). After synchronisation, they would both measure the clock rate of the opposite going proton to be running slower than their own clock, but of course when they meet again they will find exactly the same amount of time has elapsed on their respective clocks. Again, this is despite the fact they measure each other's instantaneous clock rates to be slower than their own as they pass right alongside each other. If a group of protons are equally spaced out along the circumference of the LHC ring and all circulating in the same direction, all start their clocks when they receive an omnidirectional start signal from the centre clock, they will not consider themselves to be synchronised with their immediate co-circulating neighbours. The protons ahead of them will appear to be advanced in time compared to the protons behind them. If all the counter circulating protons synchronise their clocks with the centre clock in the same manner, then they will measure the instantaneous clock rate of a nearby oncoming proton to be same as their own instantaneous clock rate (when they compare two clocks in their own frame to a single clock in going in the other direction). Whether that is a valid comparison, when they do not consider their own clocks to be synchronised is debatable. As to what the Doppler shift they will measure in the frequency of a co-rotating clock on the opposite side of the LHC I am not sure and will have to give that some more thought. Anyone got any ideas? It is not a case of reverse time dilation. This situation is very similar to the twins paradox. One twin travels away at high speed and then returns. During the journey both twins measure the clocks of the other twin to be running slower than their own clocks but when they meet again they find that the stay at home twin has aged more. It should be obvious that two clocks can not both "really" be running slower than each other and the truth is that it is difficult to say which clock is "really" running slower than the other when two observers are moving inertially relative to each other. Whenever two ideal clocks show different elapsed times, then at least one of the clocks has been moving non inertially. Comparison of clock rates of spatially separated clocks by measuring Doppler shifts of transmitted signals, shows up any asymmetry in the situation due to non inertial motion. Assuming you mean that that the protons on the dense planet are non circulating and experiencing a gravitational time dilation factor of 7500, then if you are experiencing a gravitational time dilation factor of 10, you would measure the time dilation of the protons to be 750 relative to your clocks. If you were to descend down towards the surface of the planet your time dilation would become increasingly similar to that of the protons and the time dilation of the protons relative to your clock will tend towards 7500/7500 = 1. There are quite a few tricky issues here (and a lot of questions!) and I may have got some of my answers wrong. Hopefully the other members here will pick up on my errors.
It's actually not an ellipse, at least not if you take "how it would look" to mean "how it would appear in a photo." The Lorentz contraction isn't what you actually see when you look at an object moving at relativistic speeds. You also get effects due to propagation time of the light. Turns out that a sphere still looks like a sphere, not an ellipsoid, and a circle still looks like a circle from within the plane of the circle. There's a nice video here: At the end it shows what it would look like if you were circling the earth at relativistic speed. The earth still looks like a sphere, but the different parts of the sphere are distorted in size.
What you say is not actually wrong, but what I was referring to is what what multiple co-moving observers would measure when moving in the plane of the circle. The OP stated "When I say "looks like" I mean according only to whatever relativistic transformations would apply, not what it looks like to a camera...", but then again he did add "...(but that would be another interesting question.)" so both view points are relevant. As you say, a sphere looks like sphere (from a camera point of view) at relativistic speeds just as sphere looks the same from any angle so the visual rotation that occurs has no effect on the visual appearance of a sphere if we ignore surface markings. On the other hand a circle looks like a line from the point of view of a camera moving in the plane of the circle and continues to look like a line (of the same length) at relativistic speeds. Square and rectangular boxes do appear distorted to a camera at relativistic speeds because of the lack of rotational symmetry. I think it is important to make clear that length contraction can be measured by multiple observers in the same frame piecing there local measurements together, while the view of an relativistic object from a single viewpoint such as a camera is distorted by light travel times from distant parts of the object and when those light travel times are taken into account, the length contracted shape of the object becomes apparent.
Thanks kev. I stand corrected on the combined Lorentz factor. I did the math (algebra) wrong. If I did it right this time, the Lorentz factor for two oncoming protons each with Lorentz factor k=7500 is 2*k*k-1 = 112,499,999 which agrees with what kev came up with. I did not completely follow the Doppler discussion. I will have to give some more thought to how protons observe each others clocks and perimeter clocks. I am fairly convinced (I think) that all protons observe the center clock running 7500 time faster, circumference clocks running 7500 times slower as they pass by, and that oncoming proton's see each others clocks slowed by 112.5 million. But at some time later before the next meeting, an oncoming proton's clock and circumference clock must then be observed to be running fast to make up for running slow in order to show 6ns at the next proton meeting, and likewise 12ns for the next circumference clock meeting. I would like to know where/how/why the time is made up. I still can't find much on what centrifugal force/acceleration the protons experience. Maybe their are experts here who could comment. What things look like to a camera would be interesting but I consider that another layer of complexity. For now I be happy to understand how things are observed or determined relativistically only. Another question that came to my mind is how long does a light pulse emitted forward and backward from a proton (say, inside a mirrored beam pipe) take to get back to the proton. I don't know if this would lead to any help answering my remaining questions. (By the way, from http://public.web.cern.ch/Public/en/LHC/Facts-en.html the exact LHC circumference is 26659m.) From the LHC frame of reference, the time for a forward light pulse to go around and catch up with the proton it started from should be 26659m/(c-v) = 26659m/(c*(1-sqrt(1-1/(7500*7500)))) = 10004s. From the proton's point of view the forward light pulse travels along the string (described in my first post) which is 7500 times the actual circumference. I would think the pulse travels locally along side the string at the speed of light. So the time is 7500 x 26659m / (299,792,458m/s) = 666.9ms on the proton's clock, a factor of 15000 from the LHC calculation. I was expecting 7500. What did I miss?
This might help. A while back experimenters at Brookhaven National Laboratory stored radioactive muons with a gamma (γ) of about 29.4 in a magnetic storage ring. The natural rest-frame lifetime of the muon is τ=2.2 microseconds. This is an important invariant "clock" for the muon. The circulating velocity of the muons in the ring was about βc. The observed laboratory-frame lifetime by the experimenters was γτ = 64.7 microseconds, and the average distance the muons traveled before decaying was βcγτ. Bob S
Bob S, that seems to be similar to the calculation I made from the LHC frame of reference which I think is correct. I am a little more suspicious of my second calculation.
Quick answer: The circumference of the LHC according to the proton (using its string) is 7500 times longer than the circumference according to the observers at rest with the LHC. The protons clock is running 7500 times slower than the clocks at rest with the LHC so overall the proton should measure the speed of light to be the same as measured by the observers at rest with LHC. [EDIT] I have checked your calculations and they seem to be OK. The calculations assume the speed of light is measured as being the same in both frames which addresses the issue I mentioned above. That means we are left with explaining why the difference in the time interval between the frames is twice the amount that can be accounted for by simple velocity time dilation by straight forward Special Relativity calculations. Didn't Einstein find he was out by a factor of 2, in his first draft of General Relativity when working out how much star light should be deflected by the Sun during an eclipse? I guess you would have to take into account whether the mirrors are attached to the tunnel in which case they would be moving relative to the proton or if they are attached to the string. I think it might be worth having a look at the Sagnac effect in this context too.
OK, I have now had some time to give this some more thought, I think I now know where the problem is. We both made the mistake of assuming that the proton would measure the speed of light to be c (which is normally a reasonable assumption). The time measured by proton would in fact be the time measured by the LHC frame divided by 7500. This makes the effective speed of light relative to the string equal to about 2c. If you work out the time for a pulse to go in the opposite direction and return to the proton as timed by the proton using 26659/(c+v)/7500, the speed of light according to the proton is about 4,500,000,000c! This relates to the fact that the speed of light is not constant over large distances in a non inertial frame and also to the fact that protons circulating around the LHC can not sensibly synchronise their clocks in the way that observers in an inertial frame can. Hope that helps.