LOG#015. Time of flight.

Suppose we get a beam made of massive particles. The rest mass (the names invariant mass or proper mass are also popular) is $m$ . The particle travels a distance L in its inertial frame. The particle has an energy E in that frame. Therefore, the so-called “time of flight” from two points at x=0 and x=L will be:

$T=\dfrac{L}{v}=\dfrac{m \gamma L}{m \gamma v}=\dfrac{ m \gamma L}{p}=\dfrac{m \gamma c^2 L}{pc^2}$

where we use the relativistic definition of momentum $p=m \gamma v$ . Now, knowing that the relativistic energy is $E=m \gamma c^2$ , using the time that a massless light beam uses to travel the proper distance L, easily calculated to be $\tau = L/c$ , we will get:

$T=\dfrac{E L}{pc^2}=\dfrac{E \tau}{pc}=\dfrac{\tau}{pc/E}$

But the dispersion relation in Special Relativity is a function $E=E(p,m)$ , and the squared of the relativistic momentum or momenergy yields the well known relationship

$p^2c^2 = E^2-m^2c^4$

i.e.,

$\dfrac{p^2c^2}{E^2}=1-\dfrac{m^2c^4}{E^2}$

and then,

$\dfrac{pc}{E}=\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}$

Thus, the time of flight is finally written as follows:

$\boxed{T=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}}$

This equation is very important in practical applications. Specially in Astrophysics and baseline beam experiments, like those involving the neutrinos! Indeed, we usually calculate the difference between the photon (or any other massless) time of arrival and that of massive particles, e.g. the neutrinos. Several neutrino experiments can measure this difference using a well designed experimental set-up. The difference between those times of flight (the neutrino time of flight minus the photon time of flight) is:

$T-\tau=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-\tau$

or equivalently

$\Delta t=T-\tau=\tau \left( \dfrac{1}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-1\right)$

or as well

$\displaystyle{ \boxed{Q=\dfrac{\Delta t}{\tau}= \dfrac{1}{\sqrt{1-\left( \frac{mc^2}{E}\right)^2}}-1}}$

This last expression can also be expressed in terms of the speed of light and neutrinos, since $c=L/\tau$ and $v_\nu=L/T$ , so

$\dfrac{ T/L-\tau/L}{\tau/L}=\dfrac{1/v_\nu-1/c}{1/c}=\dfrac{c-v_\nu}{v_\nu}$

Therefore,

$\boxed{Q=\dfrac{\Delta t}{\tau}=\dfrac{c-v_\nu}{v_\nu}}$

In the case of know light left-handed neutrinos, the rest mass is tiny (likely sub-eV and next to the meV scale), and then we can make a Taylor expansion for Q ( with $x=(mc/E)<<1$ ):

$Q\approx \dfrac{x^2}{2}+\mathcal{O}(x^4)$

Then, we would expect, accordingly to special relativity, of course, that

$\dfrac{c-v_\nu}{v_\nu}=\dfrac{T-\tau}{\tau}= \dfrac{1}{2}\left(\dfrac{m_\nu c^2}{E}\right)^2$

We can guess how large it is plugging “typical” values for the neutrino mass and energy. For instance, taking $l m_\nu \sim meV$ and $E\sim GeV$ , the Q value is about

$Q\sim \left( meV/GeV\right)^2= 10^{-24}$

Nowadays, we have no clock with this precision, so the neutrino mass measurement using this approach is impossible with current technology. However, it is clear that if we could make clocks with that precision, we would measure the neutrino mass with this “time of flight” procedure. It is a challenge. We can not do that in these times (circa 2012), and thus we don’t measure any time delay in baseline experiments. Then, neutrinos move with $v_\nu \approx c$ and since there is no observed delay (beyond the OPERA result, already corrected), neutrinos are, thus, ultra-relativistic particles, and for them $E=pc$ with great accuracy.

Moreover, from supernova SN1987A we do observe a delay due to the fact that stellar models predict that neutrinos are emitted before the star explodes! During the collapse, and without going deeper in technical details on the nuclear physics of dense matter, neutrinos scape almost instantaneously (in about some seconds) from the collapsing star but the photons find theirselves blocked during some hours due to the pressure of the dense matter. After some time passes, the block-out ends and photons are released. Historically, this picture was not the physical portrait of collapsing stars around the eighties of the past century, and the observations of supernovae entered, crucially, into scene in order to solve the puzzle of why neutrinos arrived before the photons (indeed, superluminal neutrinos were already tried and postulated to solve that paradox). However, SN1987A was helpful and useful due to the insight it offered about the nontrivial subject of collapsing stars and dense nuclear matter. Before SN1987A there were some assumptions about the nuclear processes inside collapsing stars that proved to be wrong after SN1987A. The puzzle was solved soon with the aid of SN1987A data taken by experimental set-ups like (Super)Kamiokande, and the interaction and collaboration between nuclear physicists, astrophysicists and high energy physicists. Generally speaking, that was the path towards the currently accepted models of nuclear physics in the inner shells of exploding stars.

In summary, photons find a several hours long block-out due to the dense collapsing material from some inner shells of the star, while neutrinos come out, and when the block-out ends, photons are shot into the sky (Bang! A supernova is born). The delayed time is consistent with the above expressions. The neutrinos arrive before the photons not because they are faster but due to the fact they are emitted before and they don’t find any resistence unlike photons; light is blocked by some time due the nuclear conditions of the matter inside the imploding star, but it ends after some hours. Furthermore, since neutrinos travel almost at speed of light, light generally is not able to reach them, even at astrophysical distances!

Remark (I): Generally, experimental physicists use a different formula in order to measure the time difference between the arrivals of neutrinos and photons. We will derive it easily:

$\boxed{\dfrac{c - v_\nu }{v_\nu}=\dfrac{\Delta t}{\tau}}$

This equation implies that

$\boxed{\dfrac{c - v_\nu }{c}=\dfrac{v_\nu \Delta t}{c\tau}=\dfrac{v_\nu T-v_\nu \tau}{c\tau}=\dfrac{L-L'}{L}}$

where L is the distance from CERN to Gran Sasso and L’ is the path length traveled by the neutrinos in the proper time ( $\tau$ ). Therefore, we write

$\dfrac{c- v_\nu }{c}=\dfrac{(T-\tau)}{\tau(c/v_\nu)}$

And now, since

$T=\dfrac{L}{v_\nu},\tau =\dfrac{L}{c}\longrightarrow \dfrac{T}{\tau}=\dfrac{c}{v_\nu}$

Thus,

$\dfrac{c - v_\nu}{c}=\dfrac{(T-\tau)}{\tau (T/\tau)}$

$\boxed{Q'=\dfrac{c - v_\nu }{c}=\dfrac{T-\tau}{T}=\dfrac{\Delta t}{T}}$

Obviously, it is an analogue of the previous formula Q. Q’ provides a measure of how much the propagation velocity of neutrinos differs from the speed of light. Experiments are consistent with a null result (the difference in velocity is measured to be zero). In conclusion, the neutrinos move almost at speed of light (with current experiments not being able to distinguish the tiny difference between both velocities, that of neutrino and the speed of light).

Remark (II): In the slides of OPERA-like experiments, you find the equation

$\boxed{\dfrac{v_\nu -c}{c}=\dfrac{\delta t}{TOF_c -\delta t}}$

and there, it is usually defined

$\delta t= TOF_c -TOF_\nu \longleftrightarrow TOF_\nu = TOF_c -\delta t$

There is a complete mathematical equivalence between my notation and those of experimental HEP neutrino talks handling the time of flight, if you note that:

$\delta t=-\Delta t \;\;\;\;\;\;\;\;\;\;\;\;\;TOF_c = \tau \;\;\;\;\;\;\;\;\;\;\;\;\; TOF_\nu=T$

Then, all the mathematical equations match.

Remark (III): An alternative way to calculate the relationship between neutrino speed and photon speed (in vacuum) is given by the next arguments. The total relativistic energy for the neutrino reads

$E_\nu=\dfrac{m_\nu c^2}{\sqrt{1-\dfrac{v_\nu^2}{c^2}}}$

Then, algebra allows us to write

$E^2=\dfrac{(m_\nu c^2)^2}{1-\dfrac{v_\nu^2}{c^2}}$

$E^2-(m_\nu c^2)^2=\dfrac{v_\nu^2 E^2}{c^2}$

$c^2E^2-c^2(m_\nu c^2)^2=v_\nu^2 E^2$

$v^2_\nu=c^2\left(\dfrac{c^2E^2-(m_\nu c^2)^2}{E^2}\right)=c^2\left( 1-\left(\dfrac{m_\nu c^2}{E}\right)^2\right)$

$v_\nu=\pm c\sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}$

That is,

$\boxed{\dfrac{\vert v_\nu -c \vert}{c}=\bigg| \sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}-1\bigg|}$

In the case of neutrinos, ultra-relativistic particles, and with high energy E compared with their rest mass, we get

$\dfrac{\vert v_\nu -c \vert}{c}\approx \dfrac{1}{2}\left(\dfrac{mc^2}{E}\right)^2$

as we obtained before.

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PS: This article was semifinalist in the 3quarksdaily 2012 online contest. Here the logo picture I got from the official semifinalist list http://www.3quarksdaily.com/3quarksdaily/2012/06/3qd-science-prize-semifinalists-2012.html

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LOG#015. Time of flight.

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