Electron with infinite mass

[Muhammad Ali]

According to Einstein, any thing moves or tends to move with or nearly equaly to the velocity of light will have its mass equal to infinity.
Therefore my question is that:
what is the reason that electrons being matter and moving with velocity nearly equal to light doesnot have mass equal to light?

[alexepascual]

Most of the time electrons move slowly when compared to the speed of light.
I think a case in which the high velocity has forced physicists to use relativistic expressions has been that of cosmic rays.
Of course also in the case of particle accelerators you might have electrons moving "fast". (just my guess, I don't know enough about particle physics)

[jtbell]

Of course also in the case of particle accelerators you might have electrons moving "fast".

The Stanford Linear Accelerator (SLAC) accelerates electrons to an energy of 50 GeV which corresponds to a speed of 0.99999999995c. (Assuming I counted the decimal places correctly! :uhh:)

[chrismuktar]

That's it's effective kinematic mass, not it's gravitational mass. All it means is that the faster something goes, the harder it is to accelerate further, and becoming infinite at the speed of light, this makes this an unreachable limit (you would have to push infinitely hard).

The mass is defined by

m=gamma * restmass

where gamme = 1/sqrt(1-(v/c)^2)

where v is the velocity of the electron and c is the speed of light.

Hope this helps,
Chris.

[alexepascual]

Aren't always gravitational and kinematic mass the same as far as we know?
I think the only thing you can say remains the same is the rest mass. The relativistic mass will increase as speed increases.
Am I wrong to say that the electrons in SLAC are attracted much more strongly to the earth than a non-relativistic electron?

[clive]

With the speed given by jtbell (v=0.9...95*c) I obtained gamma=31623. So the relativistic mass of these electrons is not very big (roughly a Fluorine nucleus at rest).

[dextercioby]

With the speed given by jtbell (v=0.9...95*c) I obtained gamma=31623. So the relativistic mass of these electrons is not very big (roughly a Fluorine nucleus at rest).

That's a really poor analogy... :yuck: A fluorine nucleus has a rest mass of approximately 20GeV (actually under 19),which is waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay below 50GeV...

Learn to manipulate numbers and physical constants...

Daniel.

[dextercioby]

That's it's effective kinematic mass, not it's gravitational mass.

Why did u bring gravity into discussion...??I'm sure u're well aware that the gravity effects (wrt relativistic mass) inside an accelerator of (fundamental) particles are much,very much less,than the ones SR predicts...

Daniel.

PS.Therefore it makes no point into bringing into discussion relativistic mass in gravity fields...

[clive]

GeV is used for energy. I calculated a MASS (Kg)!

[dextercioby]

Aren't always gravitational and kinematic mass the same as far as we know?


If i'm not mistaking,the gravity mass is
m_{rel.gravity}=\frac{m_{0}}{\sqrt{1+\Phi/c^{2}-\beta^{2}}}

So if u're defining
m_{rel.kinetic}=:\frac{m_{0}}{\sqrt{1-\beta^{2}}}

,i think you can draw your own conclusions.

Daniel.

[clive]

Hi Daniel,

I ask you to solve this problem:

"An electron is moving with constant velocity so that its relativistic (kinetic) mass is equal with the mass of Fluorine nucleus (at rest). Calculate the velocity of this electron."

BTW......what's so wrong with this analogy? Maybe the numbers are not so accurate but the message is very clear: even at the speed presented by jtbell, kinetic masses are still small.

[dextercioby]

BTW......what's so wrong with this analogy? Maybe the numbers are not so accurate but the message is very clear: even at the speed presented by jtbell, kinetic masses are still small.

You didn't get the point,did u...?The analogy is bad,because it specifies a wrong nuclei.You need a nuclei with A>=50 (titanium,vanadium,that area of periodic table) to get the analogy correct.
Numbers are important in physics.

What do you mean "small"??50 Gev cf.0.5MeV i'd say it's pretty big,wouldn't u say so??

Daniel.

[clive]

Daniel, can you answer my question (the first one)?

[dextercioby]

Are u making fun of of me???
\gamma\sim\frac{20GeV}{0.5MeV}=40,000. (1)

\beta=\frac{\sqrt{\gamma^{2}-1}}{\gamma}\sim \frac{\sqrt{1,599,999,999}}{40,000} (2)

v=\beta\cdot c\sim 7.5\cdot 10^{3}\sqrt{1,599,999,999}ms^{-1}


Daniel.

[clive]

Thanks Daniel!

Now you see that the answer is in fact the velocity given by jtbell. And the relativistic mass (Kg) of an an electron with that velocity is equal with the rest mass (Kg) of a Fluorine nucleus!!!

[Hans de Vries]

Thanks Daniel!

Now you see that the answer is in fact the velocity given by jtbell. And the relativistic mass (Kg) of an an electron with that velocity is equal with the rest mass (Kg) of a Fluorine nucleus!!!

0.99999999995c ==> gamma = 100,000
0.9999999995c ==> gamma = 31,622

The calculation is ok but the input is wrong :rolleyes: So many 9's.....

Regards, Hans

[dextercioby]

He,he,i knew there was something fishy with his calculations.Too many 9-s indeed.That's why i didn't extract the sq root and make the division through gamma.

Daniel.

[jtbell]

OK, here's how I calculated it... I started with

\frac {v}{c} = \frac {pc}{E}

which I used a lot when I was in particle physics once upon a time...

\frac {v}{c} = \frac {\sqrt {E^2 - (mc^2)^2}}{E} = \sqrt {1-\left(\frac {mc^2}{E}\right)^2}

Applying the binomial approximation

(1-x)^n \cong 1-nx

I got

\frac {v}{c} \cong 1 - \frac {1}{2} \left(\frac {mc^2} {E} \right)^2

Plugging in numbers,

\frac{v}{c} \cong 1 - \frac {1}{2} \left( \frac {0.0005 GeV}{50 GeV} \right)^2 = 1 - \frac {1}{2} 10^{-10} = 1 - 0.00000000005 = 0.99999999995

(with ten 9's)

[clive]

Yes, indeed there are ten 9's. The "equivalent" nucleus would be a "little" heavier, as Daniel told me before. :grumpy: