Relativistic motion and length contraction

[the farmer]

TL;DR Summary: An electron with kinetic energy of 50 MeV, such as an electron produced in a linear accelerator, can be shown to have a speed parameter of 0.999949c. A beam of such electrons moves along the axis of an evacuated tube that is 10 m long, measured in the reference frame S fixed in the lab. Imagine a second frame S', attached to an electron in the beam and moving with it.

(a) How long would this tube seem to be to an observer riding on the electron?

KE = gamma × mc^2 - mc^2
From the I make gamma the subject of the formula and get a solution which is 1.21 ×10^13
Then calculate the length contraction which is L = proper length ÷ gamma factor. The ni got an answer of 8.19 × 10^-13. I am stuck with this question because I did not use the speed = 0.999949c which I was given in the question.

 

[Orodruin]

The speed is superfluous if you know the mass of the electron and the kinetic energy. However, the problem does not provide the mass of the electron and therefore probably expects you to compute gamma from the speed alone.

 

[Ibix]

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

 

[Orodruin]

I haven't checked your numbers

OP should check their numbers though ... The electron rest mass is around 0.5 MeV/c^2 so gamma is certainly not 10^13

 

[the farmer]

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

What if I use 9.11 x 10^31

 

[Orodruin]

What if I use 9.11 x 10^31

9.11 x 10^31 what? Please use units. They are essential in physics.

 

[the farmer]

9.11 x 10^31 what? Please use units. They are essential in physics.

9.11× 10^-31 kg

 

[Orodruin]

9.11× 10^-31 kg

That is the same as 0.5 MeV/c^2 so you should get the same result. (Assuming that you do all of the unit conversions correctly.)

 

[Orodruin]

To add to that: Now would be a good moment to provide your full calculation, not only the results, but also what you put into all of your formulas.

 

[the farmer]

What do you meant by speed is superfluous

 

[the farmer]

Fine a

That's a fine approach if you know the electron mass (I haven't checked your numbers). You could calculate the gamma factor from the velocity as a check.

Fine approach but I didn't use the velocity that I was given in the question somehow I am in correct..

 

[PeroK]

What do you meant by speed is superfluous

In relativistic kinematics, energy and momentum are usually more useful quantities than speed. If you know the mass and energy of a particle, then you don't need to calculate its speed to do the calculations.

You can do the same in classical mechanics, in fact. It's just more often taught using speed and momentum, rather than energy and momentum.

 

[Orodruin]

somehow I am in correct..

You are incorrect because you did the math wrong. Hence me asking you to actually write out what you did:

To add to that: Now would be a good moment to provide your full calculation, not only the results, but also what you put into all of your formulas.

 

[the farmer]

In relativistic kinematics, energy and momentum are usually more useful quantities than speed. If you know the mass and energy of a particle, then you don't need to calculate its speed to do the calculations.

You can do the same in classical mechanics, in fact. It's just more often taught using speed and momentum, rather than energy and momentum.

So in short, you mean that I must ignore the given speed in my question?

 

[Orodruin]

So in short, you mean that I must ignore the given speed in my question?

No. You can use either the provided kinetic energy or the provided speed. They both correspond to the same gamma factor because the speed was computed given the correct electron mass.

 

[the farmer]

I will redo it now

 

[Orodruin]

KE = gamma × mc^2 - mc^2
From the I make gamma the subject of the formula and get a solution which is 1.21 ×10^13

For what it is worth: The first line is correct. The second line is not. Unless you provide more details on your calculation in between, it is impossible to say exactly where the error occurs.

 

[PeroK]

So in short, you mean that I must ignore the given speed in my question?

You can use either, but computing a gamma factor from a speed so close to $c$ is a bit messy. The gamma factor comes out more easily from the energy. In fact, that's why $MeV$ are useful units for high-energy particles and why mass is given in $MeV/c^2$. The electron mass is approx $0.51 MeV/c^2$. Note that often the $c^2$ is dropped. The gamma factor is simply the energy in $MeV$ divided by the mass in $MeV/c^2$.

 

[Orodruin]

The gamma factor is simply the energy in MeV divided by the mass in MeV/c2.

Note however that the OP is given kinetic energy.

Pet peeve: units should generally not be typeset in math font in typical LaTeX. For example, the mass unit would be ${\rm MeV}/c^2$ and be preceeded by a space to separate from the numerical value.

 

[the farmer]

Alright, you said the formula in the first line is correct right? Step 2 I make my gamma the subject of the formula let say gamma is = y, y= KE + mc^2 / mc^2 = 8×10^-12 J (9.11×10^-31 kg ) ×( 3×10^8m/s)^3 ÷ 9.11×10^-31×(3×19^8)^2 then my final answer for gamma is 98.57
From there i have to find l= L0/ y , L0 is proper length which is 10m, then L = 10m/ 98.57 = 0.10 m,
What I am confused of is that I didn't use speed of 0.999949c which I was given. Whic method should use if I want to use it for accuracy.

 

[Orodruin]

Whic method should use if I want to use it for accuracy.

Either. Given your input data has only one or at most two significant digits, the result will not be more accurate in either case.

 

[the farmer]

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

 

[the farmer]

Either. Given your input data has only one or at most two significant digits, the result will not be more accurate in either case.

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

 

[Ibix]

Okay that I know, my question is what method should use if I want to use both kinetic and speed that I was give to find that length in a)

You don't need both. The speed alone is enough; the kinetic energy and rest mass is enough. As Orodruin noted, they're probably expecting you to use the speed because they didn't give you the mass, but either eay works.

I presume there's a part (b) and (c) etcetera where you will need the KE.

 

[the farmer]

No there is no other questions, but which formula do need to use so that I can use the speed only without using kinetic energy. I feel like is this but I am not sure L = L0/y. Y i gamma factor then it will be L = 10m/ square root of 1- (0.999949)^2 where by c^2/c^2 will cancel each other then the solution is 990.16016776 that is my length for a

 

[Ibix]

990.16016776 what? And is that a sensible number of decimal places?

 

[the farmer]

990 m in 3 significant figures

 

[Ibix]

Right - agreed.

For future reference, note that we have $\LaTeX$ here, so you can write things like $\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$. You may need to refresh the page to get the renderer to kick in. If you're not familiar with $\LaTeX$ there's an introductory guide linked just below the reply box.

 

[the farmer]

Thanks so much? L =10 m ÷ 99.01601676
= 0.10099376117m
0.101m in 3 significant figures
Is this the right answer In a)

 

[Orodruin]

FWIW Anything more than two significant digits is wildly overstating accuracy here.