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Relativity at the LHC

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  • #1
bobie
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Homework Statement


This is not really homework, I am trying to understand relativity and I suppose my question is too simple to post in other forums.
I took the following data from the Cern site when they reached 7TeV:
2. Relevant data
Ek = 7 Tev (1.6926*1027 Hz)
B = 8.33 T
RLch = 4243 00 cm
Ep(m) = 0.938 GeV = 2,2685 * 1023 Hz (= 1,67 *10-24 g)

β= 0.999 999 991 c : I'm not sure how many 9 should be there. Can you show me how to find it?

Relativistic mass: can we find the mass increase only if we know v/c?
[tex]M+ = \frac{1} {\sqrt (1-β^2)}[/tex]

Has this curve got a name? is it similar to sec α?
Could someone plot the two curves together in a graph, please?

Can I find the exact value of the mass increase just dividing Ek by Ep:
M+ = 7000/0.938 = 7461 rest masses ?
 
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  • #2
vanhees71
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In modern high-energy physics the mass is always taken as the rest mass of the particle. In our case we deal with protons which have a restmass of [itex]m=938 \; \mathrm{MeV}/c^2[/itex]. The energy is related to momentum via
[tex]E=c \sqrt{m^2 c^2+p^2}.[/tex]
Note that in the runs so far the protons in each beam had an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex]. The total center-of-mass energy was [itex]\sqrt{s}=7 \; \mathrm{TeV}.[/itex] This will be doubled in the future after the just running upgrade to [itex]\sqrt{s}=14 \; \mathrm{TeV}[/itex], i.e., to an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex] for the protons in each beam.

From the given energy you can easily calculate [itex]p[/itex] and the velocity then is given by
[tex]v=c^2 p/E.[/tex]
I guess now you can figure everything out yourself :-).
 
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  • #3
bobie
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\mathrm 938 {MeV}/c^2[/itex].
to an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex] for the protons in each beam..
Thanks, vanhees,

- why is Ep divided by c2?
- my value for Ek is wrong? are the other data correct? R should be correct
- isn't there a Lorenz formula for v, like the one for M+?
- I suppose I need a lot of precision to get 0.999 999 991; before I try to solve it by myself, can you tell me at least if that value is correct?
- do you have the possibility of showing me that curve and how it is related to sec α?

- is M+ = 7461 correct?, I get same result with (1-√1-0.9999999912)-1: 0.000134-1 =7461

- if I use the formula mv/B =r => m = R*B/ c, what is B ? 82,200 G?
M+ = 424,300* 82,200/2.99*1010 = 1.162582 it is how many masses?

Thanks
 
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  • #4
bobie
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to an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex] for the protons in each beam..
Is that a typo , should it be 7 ?
 
  • #5
bobie
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Note that in the runs so far the protons in each beam had an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex]. The total center-of-mass energy was [itex]\sqrt{s}=7 \; \mathrm{TeV}.[/itex] This will be doubled in the future after the just running upgrade to [itex]\sqrt{s}=14 \; \mathrm{TeV}[/itex], i.e., to an energy of [itex]E=3.5 \; \mathrm{TeV}[/itex] for the protons in each beam..
I checked with the data but each proton seems to have 7 TeV:
[tex]E = \frac{m_0c^2}{\sqrt{1-(v/c)^2}}[/tex]
[tex]\frac{0.938 GeV}{\sqrt{1-(0.999 999 991)^2}}[/tex]
= 0.9383 GeV / 0.000134 = 7000 GeV

Where do I go wrong?
 
  • #6
vanhees71
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Ok, then obviously your velocity refers to the design energy of [itex]\sqrt{s}=14 \mathrm{TeV}[/itex], i.e., for [itex]E=7 \mathrm{TeV}[/itex] for the protons in one of the beams.
 
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  • #7
BruceW
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yeah, it's the right answer for 7TeV per beam. But last I heard, the LHC was running two 4TeV beams (not 7TeV).
 
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  • #8
bobie
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your velocity refers to the design energy of [itex]\sqrt{s}=14 \mathrm{TeV}[/itex]
yeah, it's the right answer for 7TeV per beam. But last I heard, the LHC was running two 4TeV beams (not 7TeV).
I took these data before the reached 8TeV, and probably the experiment regarded only one beam.

Can you help me now check the radius?, I think the formula is
R = mv/ qB
(1.67*10-27 Kg * 7461) * 3*108 / 1.6*10-19 * 8.22
I get 2840 m instead of 4243, where do I go wrong?
 
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  • #9
BruceW
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that is the non-relativistic equation. There is only a small change to turn it into the relativistic version. instead of using the rest mass, you need to use the total (relativistic) energy of the particle. (which will approximately be equal to its relativistic momentum at those speeds). you can derive this relativistic equation if you think about the four-force and relate the spatial part of the four-force to the Lorentz force. Remember that the Lorentz force is defined as:
[tex]\frac{d\vec{p}}{dt}[/tex]
while the four-force is usually defined as:
[tex]\frac{dP^\mu}{d\tau}[/tex]
so there is going to be a factor of gamma there (which I missed the first time that I tried to do this derivation). So be careful not to forget about this.
 
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  • #10
bobie
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There is only a small change to turn it into the relativistic version. instead of using the rest mass, you need to use the total (relativistic) energy of the particle....
so there is going to be a factor of gamma there.
Thanks BruceW, I am just a student: I do not know what factor you are referring to, I thought that the (Lorenz) factor gamma is the formula I mentioned in post #1 as M+ (E/m0c2), and E = 7TeV. I am lost!

Is it anyway a factor of 1.5? that is the difference I find.
- Can you tell me at least if R = 4243 m is the correct radius?
-can you tell me if the curve of M+ (when β varies from 0 to 1) is a secant h/c2, if we consider h= c and c1 = β?

Thanks for your time.
 
  • #11
BruceW
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yeah, you say your beam has energy 7TeV, so the total relativistic energy is 7TeV. So your equation for the radius should be: R=Ev/qBc^2 Where E is 7TeV and the c^2 is required to keep to SI units.

I am not sure what you mean when you talk about this M+ ... is it literally just ##1/\sqrt{1-\beta^2}## ? This is equal to the gamma factor I was talking about. If this is what you meant, then M+ is the relativistic mass, divided by the rest mass. Also, the relativistic mass is equal to the total relativistic energy, divided by c^2.
 
  • #12
vanhees71
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NO! The mass is the invariant mass. It's confusing to use the old concept of a relativistic mass. That's just energy (modulo a factor [itex]c^2[/itex]). Also when you have [itex]\sqrt{s}=14 \; \mathrm{GeV}[/itex] at a collider with two beams of the same particles (particles with the same invariant mass) this means that a particle in each beam has a total energy of [itex]E=7 \; \mathrm{GeV}[/itex] with the rest energy included!
 
  • #13
bobie
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Can you help me now check the radius?, I think the formula is R = m(+)v/ qB
(1.67*10-27 Kg (= M0)* 7461 ( γ = M+) * 3*108 / 1.6*10-19 * 8.22
I get 2840 m instead of 4243, where do I go wrong?
M+ stands for relativistic mass, mass increase, M0*γ ( Lorenz factor).
I think I have already multiplied m by γ , (the mass of the proton M0 by 7461), what else should I do to get the right R?
 
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  • #14
bobie
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the formula is [itex]R = \frac{γ m*v}{q*B }[/itex]
(1.67*10-27 Kg m0*[itex]\gamma[/itex] 7461) * 3*108 / 1.6*10-19q * B 8.22
I get 2840 m instead of 4243, where do I go wrong?
NO! The mass is the invariant mass. It's confusing to use the old concept of a relativistic mass.
So, what figure should I write instead of [itex]\gamma m[/itex]?
 
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  • #15
BruceW
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M+ stands for relativistic mass, mass increase, M0*γ ( Lorenz factor).
I think I have already multiplied m by γ , (the mass of the proton M0 by 7461), what else should I do to get the right R?
Oh yeah, you did already multiply by gamma. sorry, I missed that. uh, hmm. your answer for R should be correct, as far as I can see.
The equation ##R=m_0\gamma v/eB## is definitely correct. (and it is in SI units). And this is the equation you used... Are you certain the energy of each beam is 7TeV and the magnetic field is 8.33T ? Using these values, I get roughly the same answer as you do, and since I'm pretty sure the formula is true, I am suspicious of the values.
 
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  • #16
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OK, so R is definitely around 2800 m, probably I made a typo when I transcribed the data. Thanks.

Can you tell me what curve is y? I read that it is hyberbolic , is it as a function of β?
Does it describe the curve of a secant as a function of β?
 
  • #17
BruceW
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If we use ##\beta=\tanh(\phi)##, then ##\gamma=\cosh(\phi)## So we are creating a new parameterization, with the parameter ##\phi## (called the rapidity). And under this parameterization, the Lorentz factor is a hyperbolic function of the rapidity. I don't know about secant... that's not a hyperbolic function. So I would guess it would not be useful.

Also, about R - I also read on wikipedia that when they get the beam energy to 7TeV, they will need a magnetic field of 8.3T and they say the radius of the LHC is 4297m ... So I guess maybe your data was correct. But then I don't understand why the formula doesn't work... uh... Maybe we are missing some other piece of information.
 
  • #18
BruceW
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Ah yeah, here it is: http://www.lhc-closer.es/1/4/2/0 So it says the magnetic field is not the same all the way around the LHC, which is why we get the wrong answer when we assume there is the same magnetic field all the way around.
 
  • #19
bobie
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it says the magnetic field is not the same all the way around the LHC, which is why we get the wrong answer when we assume there is the same magnetic field all the way around.
Great!
If we use ##\beta=\tanh(\phi)##, then ##\gamma=\cosh(\phi)##
I don't know about secant... that's not a hyperbolic function. So I would guess it would not be useful.
Is it possible to explain in simple concepts what that means? and what is rapidity?

If we set h= c, c1 = β , and [itex]\lambda[/itex] the angle opposite c2, then c2 = [tex]\sqrt{1- β^2}[/tex] and
λ1 = f(β) = h/c2 = sec λ (when β goes from 0 to 1.)
Is that correct?
What is the relation of your γ (##cosh(\phi##)) to my γ1?
 
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  • #20
BruceW
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ah, you've lost me. I don't know what any of h,c,c2,##\lambda##,##\lambda_1## are meant to be.
In the equations I wrote down, ##\cosh(\phi)## does not mean the cosine of ##h\phi##. I was using ##\cosh## to be the hyperbolic cosine. So you should only need one parameter. Also, I don't understand what you mean when you start talking about 'angles opposite to'. The hyperbolic functions are not really related to triangles... unless they are triangles with imaginary angles maybe.
 
  • #21
bobie
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ah, you've lost me. I don't know what... The hyperbolic functions are not really related to triangles.
I regret I did not make myself clear:
If we want to define the curve of [itex]\gamma[/itex] as a function we can choose only [itex]\beta[/itex], (there is no other parameter in the formula):
[itex]\gamma[/itex] = f([itex]\beta[/itex]) (0 < [itex]\beta[/itex] < 1),
now, the simplest possible representation is as follows:

let's consider the right triangle (here:) ABC in which the hypothenuse h = c (speed of light) and leg a = β [a is opposite to angle λ at [itex]\hat{A}[/itex] (a= sin λ)], then leg b = √1-β2 (= cos [itex]\lambda[/itex] (A)).
Is it clear so far, is this a possible interpretation?, if so, we can rewrite our formula:
[tex]γ = \frac{1} {\sqrt (1-β^2)}[/tex]
[tex]γ = \frac{1} {cos λ}[/tex]
[itex]\gamma[/itex] = (h/b) = sec [itex]\lambda[/itex] (A).

- Is there anything wrong with this description?,
- could you explain how we can set [itex]\beta[/itex] = tan [itex]\Phi[/itex] when [itex]\beta[/itex] varies from 0 to 1 whereas the function tan varies from 0 to ∞?
 
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  • #22
BruceW
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hmm. I'm still not certain what you mean. Do you mean like if we have a triangle, with length of hypotenuse = 1 and length of one of the other sides = ##\beta## then the length of the other side must be ##\sqrt{1-\beta^2}## ? Yeah I guess that would work. I don't see any problem with it. But we would also have to specify that ##\beta## can only go from zero to 1.

When we use the hyperbolic description ##\beta = \tanh(\phi)## this forces ##\beta## to be within -1 and 1. So it's 'better' in that sense.

edit: or, in your parameterization, if we have ##\beta =\sin(\theta)## and ##\gamma=\sec(\theta)## then we should specify that ##\theta## is between zero and ##\pi /2## so that we get a unique mapping.
 
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  • #23
bobie
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Do you mean like if we have a triangle, .....
When we use the hyperbolic description ##\beta = \tanh(\phi)## this forces ##\beta## to be within -1 and 1. So it's 'better' in that sense..
Why better? β can't be <0!
What is the purpose or the reason for choosing tan insted of
What is the practical advantages of that function against sec λ?

If we plot the 2 curves, shouldn't we get same values?
Could you make a graph or give me a link where I can see the curve? I read that it looks like a quarter circle?

Thanks, Bruce, your help is invaluable!

P.S. isn't it enough if we say: f(β), 0<β<1 ?
 
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  • #24
vanhees71
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The natural way to describe the special-relativistic spacetime, the socalled Minkowski space, is to use a four-dimensional real vector space with the pseudo-scalar product
[tex]x \cdot y=\eta_{\mu \nu} x^{\mu} x^{\nu},[/tex]
where the indices run from 0 to 3 and [itex](\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)[/itex]. The 0 component denotes the time of an event (modulo a factor [itex]c[/itex], i.e., [itex]x^0=c t_x[/itex]).

To change from one inertial frame to another, moving with speed [itex]v[/itex] in [itex]x[/itex] direction is a linear transformation
[tex]x'^{\mu}={\Lambda^{\mu}}_{\nu} y^{\nu},[/tex]
such that
[tex]x'^{\mu} y'^{\nu} \eta_{\mu \nu}=x^{\mu} y^{\nu} \eta_{\mu \nu}.[/tex]
Working out the constraints of the matrix, given that the [itex]y[/itex] and [itex]z[/itex] components of the four vectors do not change (and also the sign of the time component doesn't change) gives the Lorentz transformation in the form
[tex]c t'=\cosh \eta c t-\sinh \eta x, \quad x'=\cosh \eta x-\sinh \eta c t.[/tex]
The velocity of the origin of the primed system, [itex]x'=0[/itex] thus is
[tex]v=c \tanh \eta.[/tex]
Here, [itex]\eta \in \mathbb{R}[/itex] implies that [itex]\beta=v/c=\tanh \eta[/itex] obeys [itex]|\beta|<1[/itex], i.e., the relative velocity of physically realizable inertial frames is necessarily less than the speed of light.

You can express the Lorentz transform easily in terms of [itex]\beta[/itex], which leads to the more familiar form
[tex]c t'=\gamma (c t-\beta x), \quad x'=\gamma (x-v t) \quad \text{with} \quad \gamma=\cosh \eta=\frac{1}{\sqrt{1-\beta^2}}.[/tex]
 
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  • #25
bobie
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[tex]v=c \tanh \eta.[/tex]
Here, [itex]\eta \in \mathbb{R}[/itex] implies that [itex]\beta=v/c=\tanh \eta[/itex] obeys [itex]|\beta|<1[/itex], i.e., the relative velocity of physically realizable inertial frames is necessarily less than the speed of light.
THat is very clear, thanks!
 
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