Determining who measures Proper Time and Length

[ja_tech]

It has REALLY been bugging me lately. I'm studying Einstein's Special Rel and am finding it annoying to determine who observes the proper length and time in the following conditions:

An electron accelerator is 4km long and can accelerate electrons up to speeds of .9999995c.

So when asked to find the "time of travel for the electron in the reference frame of the scientists at the accelerator lab", one would assume that this is referring to the relative time, and not the proper time.

The proper time would be the time of travel in the electron's reference frame, would it not?!!!
t=d/s
=4000/.9999995 X (3.0 X 10^8)

According to the answers in this NEAP study guide I am misguided.

Please inform me!

[vanhees71]

Proper time is always defined for a given point. It's the time measured in the reference frame where the particle is at rest (the co-moving frame).

The proper time of an Electron is the time measured in the co-moving frame. Fortunately, you do not need to do the Lorentz transformation from the lab frame (where the scientists measure time) to the co-moving frame since the proper time is a Lorentz invariant since it is defined with respect to a physically well defined frame, namely the co-moving frame. It is given by

\tau=\int_{0}^t \mathrm{d} t' \sqrt{1-v^2(t')/c^2},

where v(t) is the speed of the electron as measured in the lab frame. For an electron at constant speed this gives the relation between lab time and proper time to be

\tau=t \sqrt{1-v^2/c^2}.

[ja_tech]

... Fortunately, you do not need to do the Lorentz transformation from the lab frame (where the scientists measure time) to the co-moving frame since the proper time is a Lorentz invariant since it is defined with respect to a physically well defined frame, namely the co-moving frame. It is given by

\tau=\int_{0}^t \mathrm{d} t' \sqrt{1-v^2(t')/c^2},

where v(t) is the speed of the electron as measured in the lab frame. For an electron at constant speed this gives the relation between lab time and proper time to be

\tau=t \sqrt{1-v^2/c^2}.

I'm sorry but what exactly does it mean with 'repect to a physically well defined frame?"

[vanhees71]

It means that you define quantities with respect to special reference frames according to the physical situation at hand. One example is the proper time, which is a scalar measure of time for a given particle, dependent on its trajectory. Physically it's defined as follows:

The trajectory is given in any inertial reference frame (let's call it the lab frame) by the four vector,

(x^{\mu})=\begin{pmatrix}t \\ \vec{x}(t) \end{pmatrix}.

Here, and in the following, I set the velocity of light, c=1 (natural units).

Now you define the infinitesimal proper-time increment \mathrm{d} \tau, as the time in a momentary inertial reference frame, with respect to which the particle is at rest. This you do at any time, t, and add up all these time increments. This defines proper time.

Now the quantitiy

\mathrm{d} x^{\mu} \mathrm{d} x_{\mu}=\mathrm{d} t^2-\mathrm{d} \vec{x}^2=\mathrm{d} t^2 [1-\vec{v}(t)^2] \quad \text{with} \quad \vec{v}(t)=\frac{\mathrm{d}}{\mathrm{d} t}\vec{x}(t)

is a Lorentz invariant. For a massive particle that's always a positive quantity, and thus you can take the square root of it. Since in the momentary rest frame of the particle, you have \vec{v}=0, the proper-time increment can be expressed entirely in terms of the lab-frame-time increment via

\mathrm{d} \tau= \mathrm{d}t \sqrt{[1-\vec{v}^2(t)]}.

"Adding" these increments up thus means to integrate from the initial time, t=0, to some lab-frame time t. Thus you get the proper time of the particle as a function of lab time by the integral,

\tau(t)=\int_0^t \mathrm{d} t' \sqrt{1-\vec{v}^2(t')}.

This is a strictly growing function, and thus you can use \tau as well as a measure of time as the lab time, t. The good thing is that the proper time, by construction, is a Lorentz scalar.

This gives you the opertunity to describe the motion of the particle in a manifestly Lorentz covariant by giving the lab-frame coordinates as function of proper rather than lab-frame time. Then you describe the motion as a trajectory in four-dimensional space time, x^{\mu}(\tau) with a scalar parameter, which is "natural" in the sense that it is physically determined by the situation at hand, namely the motion of a single particle.

Then the four-velocity is given by

u^{\mu}(\tau)=\frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu}(\tau),

which is a Lorentz-covariant four-vector as is x^{\mu} since the proper time, \tau is a scalar.

Of course, you always have

u^{\mu}(\tau) u_{\mu}(\tau)=\left (\frac{\mathrm{d} t}{\mathrm{d} \tau} \right)^2 - \left (\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \right)^2=1.

then the energy-momentum four vector of the particle is defined by

p^{\mu}=m u^{\mu},

where m is the proper (or invariant) mass of the particle, which is a scalar. Because of the previous equation, you have the relation

p_{\mu} p^{\mu}=E^2-\vec{p}^2=m^2 \; \Rightarrow \; E=\sqrt{\vec{p}^2+m^2}.

This manifestly covariant description is very useful to find Lorentz covariant equations of motion.

[Doc Al]

It has REALLY been bugging me lately. I'm studying Einstein's Special Rel and am finding it annoying to determine who observes the proper length and time in the following conditions:

An electron accelerator is 4km long and can accelerate electrons up to speeds of .9999995c.
The accelerator is 4 km long in the laboratory frame. (That distance is contracted from the electron's frame.)

So when asked to find the "time of travel for the electron in the reference frame of the scientists at the accelerator lab", one would assume that this is referring to the relative time, and not the proper time.

The proper time would be the time of travel in the electron's reference frame, would it not?!!!
Yes.
t=d/s
=4000/.9999995 X (3.0 X 10^8)
This gives the correct answer to the question you quoted. (The travel time according to the laboratory observers.)

According to the answers in this NEAP study guide I am misguided.
What does the study guide say?

[ghwellsjr]

Make sure your are doing this calculation:

4000/[.9999995 X (3.0 X 10^8)] = .000013333334

and not this calculation:

[4000/.9999995] X (3.0 X 10^8) = 1200000600000.3