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I enquired after the effect of gravity at relativisic speed and, though some comments simply suggested that "...is resolved with the velocity addition formula just like all the others", the excellent answer by John Rennie showed that it is way too complicated to find, since GR is involved.

That question was closed as off topic , probably since I posed a concrete problem (setting $v$ at $0.866 c$ and acceleration at $1 \text{m}/\text{s}^2$) and someone thought it was a HW question. I'll not make the same mistake now and simply ask what happens to the electric acceleration when a charge is travelling at near $c$: is just the velocity addition formula enough to calculate the effective increase of vlocity, since GR is not involved. If not, what is the formula I need.

If you have some data about concrete experiments, please give some or a link, I suppose that, unlike gravity, these situation are examined daily in details at places like LHC

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  • $\begingroup$ what is the velocity that is added? to what? see this hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel.html $\endgroup$
    – anna v
    Mar 15 '17 at 6:38
  • $\begingroup$ Are you saying that we can apply the velocity addition formula here? so, if the charge is going at .866c and 1m/s^2 is added, the effective increase is 25 cm/s? Is the outcome different in the case of gravity? shouldn't it be the same? $\endgroup$
    – user137879
    Mar 15 '17 at 8:28
  • $\begingroup$ yes , but you should know what you are adding. look at the link, there are three velocities and two observers, $\endgroup$
    – anna v
    Mar 15 '17 at 9:38
  • $\begingroup$ Thanks, anna, could you write an answer using any example you like? Can you retrieve some data of experiments ? At LHC they accelerate charges up to .999999991 c so they should have some precise data at various speedsand probably a curve of the relativistic increase from .1 c? Can you explain why if the pull is on the mass (gravity) and not on the charge the result should be sensibly different? $\endgroup$
    – user137879
    Mar 15 '17 at 11:45
  • $\begingroup$ At LHC they use the relativistic formulas they do not have to gather data of the type you are asking. And the experiments are analyzed with four vectors which are lorenz covariant. Velocities would be confusing the issue as far as particle physics goes. You should study the links in the link above which have the example hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c1 . It should work no matter what powers the velocity, gravity or electromagnetsm. $\endgroup$
    – anna v
    Mar 15 '17 at 11:53
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This can be done very easily without worrying about relativistic addition of velocities because if a charge $q$ crosses a potential difference $V$ then its energy changes by $\Delta E = qV$.

A quick aside: the LHC actually uses RF cavities to accelerate the protons, but let's leave aside that complication and just assume the charged particle repeatedly crosses a potential difference $V$.

The energy of a relativistic particle is:

$$ E^2 = p^2c^2 + m^2c^4 \tag{1} $$

where $p$ is the relativistic momentum:

$$ p = \gamma mv = \frac{mv}{\sqrt{1 - v^2/c^2}} $$

and with a bit of algebra we can rearrange this into an expression for the velocity in terms of the energy:

$$ v = c\,\sqrt{1 - \frac{m^2c^4}{E^2}} \tag{2} $$

So if we start with some energy $E$ and cross a potential difference $V$ the energy changes from $E$ to $E+qV$. So if our initial velocity is:

$$ u = c\,\sqrt{1 - \frac{m^2c^4}{E^2}} $$

then the final velocity is:

$$ v = c\,\sqrt{1 - \frac{m^2c^4}{(E+qV)^2}} $$

And if this occurs over some short distance $\ell$ we can approximate the acceleration using the SUVAT equation:

$$ v^2 = u^2 + 2as $$

to get:

$$ a = \frac{c^2}{2\ell} \left(1 - \frac{m^2c^4}{(E+qV)^2} - 1 + \frac{m^2c^4}{E^2}\right) $$

which for reasons that will shortly become clear I'm going to rearrange to:

$$ a = \frac{m^2c^6}{2\ell} \left( \frac{\frac{qV}{E}\left(2 + \frac{qV}{E}\right)}{E^2\left(1+\frac{qV}{E}\right)^2} \right) \tag{3} $$

You could be forgiven for wondering where on Earth all this algebra is going, but I'm now going to make the assumption that the energy gained in crossing our electric field, $qV$, is much less than the total energy $E$. This is reasonable even for a stationary particle because remember that $E$ includes the rest mass energy so its minimum value is $E+mc^2$ even when the particle is stationary. Since with this assumption $qV/E \ll 1$ our equation (3) simplifies drastically to:

$$ a \approx \frac{m^3c^6}{E^3} \left( \frac{qV}{m\ell} \right) $$

Suppose our particle is stationary then the energy is $E=mc^2$ and our equation becomes:

$$ a_0 = \frac{qV}{m\ell} $$

And this is just the classical acceleration of a charge $q$ in a field gradient $V/\ell$. So we can write our final equation as:

$$ a = \frac{m^3c^6}{E^3} a_0 \tag{4} $$

where $a_0$ is the acceleration at non-relativistic speeds. And there's our result. The acceleration we observe in the lab is less than the acceleration at non-relativistic speeds by a factor of $m^3c^6/E^3$.

If you haven't completely lost the will to live by now there is one more step we can take. For very highly relativistic particles the energy is much greater than the rest mass and our original energy equation (1) becomes:

$$ E\approx pc \approx \gamma mv$$

Substituting this in our equation (4) we get:

$$ \frac{a}{a_0} \approx \frac{c^3}{\gamma^3v^3} $$

And since for highly relativistic particles $v \approx c$ we end up with:

$$ \frac{a}{a_0} \approx \frac{1}{\gamma^3} \tag{5} $$

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  • $\begingroup$ Thanks a lot, can we apply the same trick of potential difference in a gravity field? anna says (above) that the result is the same: "*... It should work no matter what powers the velocity, gravity or electromagnetism. *"If she is right, why in the other question you didn't take such a shortcut and used GR? $\endgroup$
    – user137879
    Mar 16 '17 at 6:10
  • $\begingroup$ @lambertwhite: The trick doesn't work in GR. It works here because spacetime is flat and we need only consider the effects of velocity, which vastly simplifies the problem. $\endgroup$ Mar 16 '17 at 6:17
  • $\begingroup$ I see, thanks, so anna and dmckee are wrong. I hope you will spare some time, sometime, and compare the effects of 1m of acceleration due to gravity and electrostatics at any speed you like, just to give a rough idea of the difference. Here, the formula for v = .866 c is $1/\gamma^2$ $\endgroup$
    – user137879
    Mar 16 '17 at 7:05
  • $\begingroup$ @lambertwhite: when an object is falling into a black hole we have gravitational time dilation as well as the effects of velocity. The object first accelerates inwards then decelerates to a halt at the event horizon. The behaviour is totally different. The behaviour will be similar to the electrostatic case above if, and only if, the gravitational fields are everywhere weak. $\endgroup$ Mar 16 '17 at 7:10

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