9 ms·

Does the LHC collide protons at twice the speed of light?

128 points by jacobedawson 8y ago

ymolodtsov 8y ago
No.
lisper 8y ago
Heh, I just now published a blog post of my own about this same topic: http://blog.rongarret.info/2019/04/we-interrupt-this-blog-to-bring-you.html http://blog.rongarret.info/2019/04/we-interrupt-this-blog-to...
Gibbon1 8y ago
I think it would worth pointing out that the two protons smash into each other with a lot more energy than you would think if you thought kinetic energy was proportional to speed squared.
- Gibbon1 8y ago
  Coming back to link to this. Indicative of the amount of energy being pumped into the beam. And this is 'just' SLAC. https://today.slac.stanford.edu/feature/beam-dump.asp https://today.slac.stanford.edu/feature/beam-dump.asp
cs02rm0 8y ago
This is the same physics I was taught in school as for two cars crashing into each other at 70mph I presume? Neither has a 140mph impact, although a 140mph impact would involve more than double the energy anyway AIUI.
- archgoon 8y ago
  No, not unless you were making relativistic corrections for the vehicles, which would be silly[1]. From the perspective of either vehicle, you do see an oncoming vehicle traveling at (approximately) 140 miles per hour. When dealing with things traveling close to the speed of light, this doesn't work, because the relativistic correction term becomes significant. The equation they show on the blog (u + v)/(1+uv) represents the velocities in fractions of speed of light. When u and v are much smaller than the speed of light, the denominator can be ignored, as it's close to 1, and we simply get (u+v). [1] But still pedagogically sound when teaching relativity :)
  - xxxvvv 8y ago
    For the people who (like myself) are confused by the missing c^2 in the blog, (u + v)/(1+(uv/(c^2)) is the original. Much easier to understand that the "simplified" version, which is usually the case.
    - zinclozenge 8y ago
      In relativity, c=1 is a common convention. It doesn't really add much to the equation because it's just a conversion factor.
      - archgoon 8y ago
        Omitting conversion factors is a bit like omitting type information. If you already know what everything is, it doesn't add information, and can make the code look simpler. If you don't, not omitting them can help make sense of things when you're learning and lack context.
      - wzcard 8y ago
        To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that contains c^2. Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at: c * (a * b) / (1 + a * b) Then you convert the car speeds to fractions, only to multiply back by c=299792458. Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.
- maxander 8y ago
  No, I don't think so- two cars impacting head-on at 70mph would have a "140mph impact," insofar as that it would be essentially equivalent to one car traveling at 140mph hitting a stationary car.
  - kbenson 8y ago
    Hmm, wouldn't a car of the same mass going the same speed joust counteract all the force exactly? If one of the cars had half as much mass, wouldn't that lessen the impact of the larger car and increase it for the smaller car, as the combined center of gravity afterwards would be moving inthe direction the larger car was traveling? It's been far too long since I had a physics class...
    - maxander 8y ago
      Well, maybe we're using different terms- as the article notes, there's technically no such thing as a "such-and-such speed impact." I was speaking in terms of how much kinetic energy is liable to go towards damaging the cars involved (or into the fireball of a particle collision); in that case, if you have X energy in two cars, or 2X energy in one and 0 in the other, the sum is the same. If we consider debris or etc flying away after the collision, then you're right that they're different- the combined velocity of the two cars is 0mph in the both-moving case and 140mph (in some direction) in the one-moving case, and the velocity of all the debris will sum to that velocity. (And the sum of velocities would be weighted appropriately if one of the cars was more massive than the other.)
    - auslander 8y ago
      On a side note, if a massive 4wd heads-on collides with Corolla, the 4wd guy will have scratches and bruises, Corolla people would not suffer, instant death. Choosing a car, choose tons, not stars :)
      - saagarjha 8y ago
        It's a sad reality that just makes the road less safe for everyone and hurts the environment :(
- saagarjha 8y ago
  Depends on your reference frame. In the reference frame of the approaching car you have a 140 mph impact.
- Waterluvian 8y ago
  That concept still feels intuitively wrong to me.
kmm 8y ago
Well in a sense yes. The distance between the two protons decreases at almost exactly twice the speed of light, as seen from the rest-frame of the detector. That's not a violation of relativity as there is no physical object or information moving that fast. And of course either proton sees the other approaching at just a little less than c
- iamgopal 8y ago
  Explain your last sentence ?
  - saagarjha 8y ago
    From the frame of reference of each proton, the opposing proton is going nearly the speed of light. Edit: someone asked for a more detailed explanation, so here it is. Because of the equation in the article, it's impossible to see anything as moving faster than the speed of light in any reference frame; they just appear to asymptotically approach it. Both protons are moving at nearly the speed of light and perceive the other as moving at a speed that's slightly closer to the speed of light.
    - deleted 8y ago
      [deleted]
  - AceJohnny2 8y ago
    As an entity approaches the speed of light, its "timeframe" slows down. So the timeframe of one proton is slowed down so much that, to it, the speed of the incoming proton is only the speed of light. A key insight of Einstein's relativity is that the speed of time isn't fixed and absolute. Hence, "relativity". I highly recommend Chris Lee's excellent 2012 article on the history of "the speed of light" at Ars Technica: https://arstechnica.com/science/2012/09/the-complicated-truth-behind-scientific-findings/ https://arstechnica.com/science/2012/09/the-complicated-trut...
    - jujulet 8y ago
revskill 8y ago
If photon travels at speed of light, then it has no mass ? But if it has no mass, it shouldn't bring any energy, though.
- 781 8y ago
  https://www.quora.com/How-can-photons-have-no-mass-and-yet-still-have-energy-given-that-E-mc-2 https://www.quora.com/How-can-photons-have-no-mass-and-yet-s...
- gus_massa 8y ago
  The protons in this device travel almost at the speed of light.
ajkjk 8y ago
Intuition for relativistic velocity works a lot better if you imagine changing units so that velocities between (0,c) become velocities in (0, infinity). Then two infinities added is just infinity, and adding two numbers which are close to infinity is a number even closer to infinity. This amounts to measuring 'rapidity' w (https://en.wikipedia.org/wiki/Rapidity https://en.wikipedia.org/wiki/Rapidity) instead of velocity. Rapidity captures what velocity 'actually is': a hyperbolic angle between the spatial coordinates and the time axis, given by w = arctanh(v/c). For low velocities, w = v/c, so you can write v = wc and just treat it like a regular velocity and it basically adds as you'd expect. But as v/c -> 1, w -> infinity. (Unfortunately, non-trivial rapidities along different spatial directions don't exactly add, and it's complicated. Boosting in x and y can result in a rotation in the xy plane also. Or something like that; I haven't looked at the math in a while.) I sometimes wonder if it would make more sense to define cw = v as the 'correct' extension of velocity to high energies, and just say "oh, meters per second -- that was wrong. That's not what velocity is". Thinking of the speed of light as 'infinite' seems very appealing.
- jakeinspace 8y ago
  The small angle approximation strikes again!
- deleted 8y ago
  [deleted]
- noobermin 8y ago
  The problem of course is when position is actually important in a problem, you need to integrate velocities instead of rapidity. The reason velocity addition trips people up is because adding velocities inherently means "seeing velocity as measured by another frame," which then requires Lorentz transform. For many problems, you confine yourself to a single inertial frame, you can still do physics comfortably and you never have to worry about adding velocities.
- tzs 8y ago
  Speaking of special relativity, people learning basic special relativity might find this simple special relativity calculator I made useful [1]. It's fairly primitive. The original intent was that it would be a base to play around with CSS on, and to maybe rewrite using various front end frameworks to learn them, and add features like named variables, showing Minkowski diagrams, cloning rows, and things like that. I never got any time for any of that, so it never got past the very simple stage, but is fine for doing things like figuring out the ladder paradox and things of that ilk when learning about special relativity. [1] https://github.com/tzs/Physics-special-relativity-calculator https://github.com/tzs/Physics-special-relativity-calculator
JshWright 8y ago
If you find these sorts of questions interesting, I can't recommend the PBS Space Time YouTube channel highly enough. https://www.youtube.com/channel/UC7_gcs09iThXybpVgjHZ_7g https://www.youtube.com/channel/UC7_gcs09iThXybpVgjHZ_7g
auslander 8y ago
> .. 1 plus 1 is not equal to 2. Which leads to contradiction. Therefore, the initial assumption, that Earth is not flat, must be false. [0] [0] https://en.m.wikipedia.org/wiki/Proof_by_contradiction https://en.m.wikipedia.org/wiki/Proof_by_contradiction
- dylan604 8y ago
  Of course 1 + 1 != 2. 1 + 1 = 10. Somebody forgot to carry the 1
p1mrx 8y ago
> you will see that adding twice 99% of the speed of light brings you to something like 99,9999% of the speed of light That seems like too many nines... why write the formula, and then not bother running the numbers through it? (.99+.99)/(1+.99*.99) is roughly 99.995%
sytelus 8y ago
Here's another way to think about this: Your velocity is time dependent quantity. When you are moving faster time for you is simultaneously slowing down. When you almost approach speed of light relative to anything, time for you slows down to almost zero relative to that thing. So everything runs in super slow motion. Another particle coming at you at the speed of light towards you will appear almost still. So you total relative velocity wrt to other particle is still just speed of light. The key is again, its because time has slowed down for you.
- gus_massa 8y ago
  This comment is confusing/wrong. > time for you slows down to almost zero relative to that thing This is at least confusing. You would not fell anything special. If all the windows are covered, you will not note that you are moving fast or slow. > Another particle coming at you at [almost] the speed of light towards you will appear almost still. If you are moving at a speed of 99% of c in the "laboratory" reference frame, and the other particle is still in the in the "laboratory" reference frame, then (if the front windshield is not covered) you will see that the other particle comes to you at a speed of 99% of c. If you are moving at a speed of 99% of c in the "laboratory" reference frame, and the other particle is moving in the oposite direction in the in the "laboratory" reference frame, then (if the front windshield is not covered) you will see that the other particle comes to you at a speed of 99.995% of c. (See the calculation details in the article.)
krick 8y ago
That's a really bad explanation IMO. The author provides some weird formula which you have to use when u and v are "near the spead of light", w/o actually explaining how near the speed of light you have to be to use (u + v) / (1 + uv) instead of u + v or even explaining why do you have to use another formula at all. She basically says "wow, look at all this, what a nonsense! but I promise it's true". Meanwhile the problem is really really simple. No, Einstein hasn't "taught us nothing travels faster than the speed of light". Instead, he taught us that neutonian velocity-addition formula is imprecise, instead of u + v it's always (u + v)/(1 + uv/c^2). It just doesn't usually matter, since "wrong" u + v formula works good enough for our everyday needs, because "c" is so huge compared to our usual "u" and "v" (so uv/c^2 is usually close to zero). And that's it, that's why velocity-addition of 0.99c and 0.99c gives us 0.9999c. Not all that "sometimes 1 + 1 equals 1" nonsense.
pontifier 8y ago
One strange thought that struck me when studying relativistic time dilation and length contraction is about light from distant stars. For an observer traveling with a photon from a distant star that reaches your eye, they would see a collision between the atom that emited the photon, and the back of your eye because there is both no space, and no time separating them.
ryanthedev 8y ago
Terrible analogy. 1+1=1. More like Ax + Bx = 1.

128 points by jacobedawson 8y ago

No.

Heh, I just now published a blog post of my own about this same topic: http://blog.rongarret.info/2019/04/we-interrupt-this-blog-to-bring-you.html http://blog.rongarret.info/2019/04/we-interrupt-this-blog-to...

Gibbon1 8y ago

I think it would worth pointing out that the two protons smash into each other with a lot more energy than you would think if you thought kinetic energy was proportional to speed squared.

Gibbon1 8y ago
Coming back to link to this. Indicative of the amount of energy being pumped into the beam. And this is 'just' SLAC. https://today.slac.stanford.edu/feature/beam-dump.asp https://today.slac.stanford.edu/feature/beam-dump.asp

cs02rm0 8y ago

This is the same physics I was taught in school as for two cars crashing into each other at 70mph I presume? Neither has a 140mph impact, although a 140mph impact would involve more than double the energy anyway AIUI.

archgoon 8y ago
No, not unless you were making relativistic corrections for the vehicles, which would be silly[1]. From the perspective of either vehicle, you do see an oncoming vehicle traveling at (approximately) 140 miles per hour. When dealing with things traveling close to the speed of light, this doesn't work, because the relativistic correction term becomes significant. The equation they show on the blog (u + v)/(1+uv) represents the velocities in fractions of speed of light. When u and v are much smaller than the speed of light, the denominator can be ignored, as it's close to 1, and we simply get (u+v). [1] But still pedagogically sound when teaching relativity :)
- xxxvvv 8y ago
  For the people who (like myself) are confused by the missing c^2 in the blog, (u + v)/(1+(uv/(c^2)) is the original. Much easier to understand that the "simplified" version, which is usually the case.
  - zinclozenge 8y ago
    In relativity, c=1 is a common convention. It doesn't really add much to the equation because it's just a conversion factor.
    - archgoon 8y ago
      Omitting conversion factors is a bit like omitting type information. If you already know what everything is, it doesn't add information, and can make the code look simpler. If you don't, not omitting them can help make sense of things when you're learning and lack context.
    - wzcard 8y ago
      To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that contains c^2. Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at: c * (a * b) / (1 + a * b) Then you convert the car speeds to fractions, only to multiply back by c=299792458. Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.
maxander 8y ago
No, I don't think so- two cars impacting head-on at 70mph would have a "140mph impact," insofar as that it would be essentially equivalent to one car traveling at 140mph hitting a stationary car.
- kbenson 8y ago
  Hmm, wouldn't a car of the same mass going the same speed joust counteract all the force exactly? If one of the cars had half as much mass, wouldn't that lessen the impact of the larger car and increase it for the smaller car, as the combined center of gravity afterwards would be moving inthe direction the larger car was traveling? It's been far too long since I had a physics class...
  - maxander 8y ago
    Well, maybe we're using different terms- as the article notes, there's technically no such thing as a "such-and-such speed impact." I was speaking in terms of how much kinetic energy is liable to go towards damaging the cars involved (or into the fireball of a particle collision); in that case, if you have X energy in two cars, or 2X energy in one and 0 in the other, the sum is the same. If we consider debris or etc flying away after the collision, then you're right that they're different- the combined velocity of the two cars is 0mph in the both-moving case and 140mph (in some direction) in the one-moving case, and the velocity of all the debris will sum to that velocity. (And the sum of velocities would be weighted appropriately if one of the cars was more massive than the other.)
  - auslander 8y ago
    On a side note, if a massive 4wd heads-on collides with Corolla, the 4wd guy will have scratches and bruises, Corolla people would not suffer, instant death. Choosing a car, choose tons, not stars :)
    - saagarjha 8y ago
      It's a sad reality that just makes the road less safe for everyone and hurts the environment :(
saagarjha 8y ago
Depends on your reference frame. In the reference frame of the approaching car you have a 140 mph impact.
Waterluvian 8y ago
That concept still feels intuitively wrong to me.

kmm 8y ago

Well in a sense yes. The distance between the two protons decreases at almost exactly twice the speed of light, as seen from the rest-frame of the detector. That's not a violation of relativity as there is no physical object or information moving that fast. And of course either proton sees the other approaching at just a little less than c

iamgopal 8y ago
Explain your last sentence ?
- saagarjha 8y ago
  From the frame of reference of each proton, the opposing proton is going nearly the speed of light. Edit: someone asked for a more detailed explanation, so here it is. Because of the equation in the article, it's impossible to see anything as moving faster than the speed of light in any reference frame; they just appear to asymptotically approach it. Both protons are moving at nearly the speed of light and perceive the other as moving at a speed that's slightly closer to the speed of light.
  - deleted 8y ago
    [deleted]
- AceJohnny2 8y ago
  As an entity approaches the speed of light, its "timeframe" slows down. So the timeframe of one proton is slowed down so much that, to it, the speed of the incoming proton is only the speed of light. A key insight of Einstein's relativity is that the speed of time isn't fixed and absolute. Hence, "relativity". I highly recommend Chris Lee's excellent 2012 article on the history of "the speed of light" at Ars Technica: https://arstechnica.com/science/2012/09/the-complicated-truth-behind-scientific-findings/ https://arstechnica.com/science/2012/09/the-complicated-trut...
  - jujulet 8y ago

revskill 8y ago

If photon travels at speed of light, then it has no mass ? But if it has no mass, it shouldn't bring any energy, though.

781 8y ago
https://www.quora.com/How-can-photons-have-no-mass-and-yet-still-have-energy-given-that-E-mc-2 https://www.quora.com/How-can-photons-have-no-mass-and-yet-s...
gus_massa 8y ago
The protons in this device travel almost at the speed of light.

ajkjk 8y ago

Intuition for relativistic velocity works a lot better if you imagine changing units so that velocities between (0,c) become velocities in (0, infinity). Then two infinities added is just infinity, and adding two numbers which are close to infinity is a number even closer to infinity. This amounts to measuring 'rapidity' w (https://en.wikipedia.org/wiki/Rapidity https://en.wikipedia.org/wiki/Rapidity) instead of velocity. Rapidity captures what velocity 'actually is': a hyperbolic angle between the spatial coordinates and the time axis, given by w = arctanh(v/c). For low velocities, w = v/c, so you can write v = wc and just treat it like a regular velocity and it basically adds as you'd expect. But as v/c -> 1, w -> infinity. (Unfortunately, non-trivial rapidities along different spatial directions don't exactly add, and it's complicated. Boosting in x and y can result in a rotation in the xy plane also. Or something like that; I haven't looked at the math in a while.) I sometimes wonder if it would make more sense to define cw = v as the 'correct' extension of velocity to high energies, and just say "oh, meters per second -- that was wrong. That's not what velocity is". Thinking of the speed of light as 'infinite' seems very appealing.

jakeinspace 8y ago
The small angle approximation strikes again!
deleted 8y ago
[deleted]
noobermin 8y ago
The problem of course is when position is actually important in a problem, you need to integrate velocities instead of rapidity. The reason velocity addition trips people up is because adding velocities inherently means "seeing velocity as measured by another frame," which then requires Lorentz transform. For many problems, you confine yourself to a single inertial frame, you can still do physics comfortably and you never have to worry about adding velocities.
tzs 8y ago
Speaking of special relativity, people learning basic special relativity might find this simple special relativity calculator I made useful [1]. It's fairly primitive. The original intent was that it would be a base to play around with CSS on, and to maybe rewrite using various front end frameworks to learn them, and add features like named variables, showing Minkowski diagrams, cloning rows, and things like that. I never got any time for any of that, so it never got past the very simple stage, but is fine for doing things like figuring out the ladder paradox and things of that ilk when learning about special relativity. [1] https://github.com/tzs/Physics-special-relativity-calculator https://github.com/tzs/Physics-special-relativity-calculator

JshWright 8y ago

If you find these sorts of questions interesting, I can't recommend the PBS Space Time YouTube channel highly enough. https://www.youtube.com/channel/UC7_gcs09iThXybpVgjHZ_7g https://www.youtube.com/channel/UC7_gcs09iThXybpVgjHZ_7g

auslander 8y ago

> .. 1 plus 1 is not equal to 2. Which leads to contradiction. Therefore, the initial assumption, that Earth is not flat, must be false. [0] [0] https://en.m.wikipedia.org/wiki/Proof_by_contradiction https://en.m.wikipedia.org/wiki/Proof_by_contradiction

dylan604 8y ago
Of course 1 + 1 != 2. 1 + 1 = 10. Somebody forgot to carry the 1

p1mrx 8y ago

> you will see that adding twice 99% of the speed of light brings you to something like 99,9999% of the speed of light That seems like too many nines... why write the formula, and then not bother running the numbers through it? (.99+.99)/(1+.99*.99) is roughly 99.995%

sytelus 8y ago

Here's another way to think about this: Your velocity is time dependent quantity. When you are moving faster time for you is simultaneously slowing down. When you almost approach speed of light relative to anything, time for you slows down to almost zero relative to that thing. So everything runs in super slow motion. Another particle coming at you at the speed of light towards you will appear almost still. So you total relative velocity wrt to other particle is still just speed of light. The key is again, its because time has slowed down for you.

gus_massa 8y ago
This comment is confusing/wrong. > time for you slows down to almost zero relative to that thing This is at least confusing. You would not fell anything special. If all the windows are covered, you will not note that you are moving fast or slow. > Another particle coming at you at [almost] the speed of light towards you will appear almost still. If you are moving at a speed of 99% of c in the "laboratory" reference frame, and the other particle is still in the in the "laboratory" reference frame, then (if the front windshield is not covered) you will see that the other particle comes to you at a speed of 99% of c. If you are moving at a speed of 99% of c in the "laboratory" reference frame, and the other particle is moving in the oposite direction in the in the "laboratory" reference frame, then (if the front windshield is not covered) you will see that the other particle comes to you at a speed of 99.995% of c. (See the calculation details in the article.)

krick 8y ago

That's a really bad explanation IMO. The author provides some weird formula which you have to use when u and v are "near the spead of light", w/o actually explaining how near the speed of light you have to be to use (u + v) / (1 + uv) instead of u + v or even explaining why do you have to use another formula at all. She basically says "wow, look at all this, what a nonsense! but I promise it's true". Meanwhile the problem is really really simple. No, Einstein hasn't "taught us nothing travels faster than the speed of light". Instead, he taught us that neutonian velocity-addition formula is imprecise, instead of u + v it's always (u + v)/(1 + uv/c^2). It just doesn't usually matter, since "wrong" u + v formula works good enough for our everyday needs, because "c" is so huge compared to our usual "u" and "v" (so uv/c^2 is usually close to zero). And that's it, that's why velocity-addition of 0.99c and 0.99c gives us 0.9999c. Not all that "sometimes 1 + 1 equals 1" nonsense.

pontifier 8y ago

One strange thought that struck me when studying relativistic time dilation and length contraction is about light from distant stars. For an observer traveling with a photon from a distant star that reaches your eye, they would see a collision between the atom that emited the photon, and the back of your eye because there is both no space, and no time separating them.

ryanthedev 8y ago

Terrible analogy. 1+1=1. More like Ax + Bx = 1.