Does a proton hurts hitting me at 0.9999 c?

[Tachyon son]

Sorry if the question is too stupid, but,
How about receiving an impact of a proton (or electron, with less mass) fired from a particle accelerator at just below c?

What will be the consecuencies in my body?

 

[chroot]

It won't cause pain, because it contains far too little energy to actually stimulate your pain receptors.

The proton will gradually lose energy by collisions with the atoms in your body. It will ionize those atoms, and leave a little streak of ionization behind it. (The same effect is exploited in bubble chambers, which allow physicists to see the tracks of particles.)

In most cases, the ionization won't really affect you at all. If the proton happens to ionize an important part of a cell, it will cause some damage. It probably wouldn't be enough damage to seriously harm the cell. If the proton happens to ionize some of the important atoms in a strand of DNA, it might eventually cause a transcription error, which, if unchecked, could result in a form of cancer.

In general, it takes thousands or millions of such particles slamming into your body to actually kill cells or cause enough DNA damage to lead to cancers. One proton really won't do much at all.

- Warren

 

[marlon]

It won't cause pain, because it contains far too little energy to actually stimulate your pain receptors.

True story

The proton will gradually lose energy by collisions with the atoms in your body. It will ionize those atoms, and leave a little streak of ionization behind it. (The same effect is exploited in bubble chambers, which allow physicists to see the tracks of particles.)

hmm, i don't think you can be too certain about these statements, though i am not saying they are wrong. For example it is quite possible that the proton will go directly through the body since there is far too little charge and mass present to trigger an actual interaction.

If the proton happens to ionize an important part of a cell, it will cause some damage.

Ionize a part of a cell ?

If the proton happens to ionize some of the important atoms in a strand of DNA, it might eventually cause a transcription error, which, if unchecked, could result in a form of cancer.

The transcription error does not need to be unchecked. Basically it is multiplied when other cells are generated yielding some "false" biological entity that we call tumor. These tumors are only evil when they use blood of the human body. Via this process the cancer can spread out over the entire body and then you are f***ed...


regards
marlon

 

[chroot]

marlon,

I considered that maybe the proton would actually go right through you, but I figured the human body, being mostly water (and therefore mostly hydrogen) would react somewhat similarly to a bubble chamber. I didn't do the calculations to determine the mean free path or anything, but I could.

And the body has a bewildering variety of mechanisms to repair damaged DNA -- cancer can only form is all of those mechanisms fail.

- Warren

 

[marlon]

Hi chroot,...

i can only agree with your words here...
I just wanted to give some extra info on possible things that might happen...
Let's not talk too much about cancer because that can't be a good thing



regards
marlon

 

[NateTG]

It's possible.

The momentum of the proton would be...
$$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$$
Now, let's say that [itex]v=c \times (1-10^{-k})[/tex]<br /> then $$\sqrt{1-\frac{v^2}{c^2}} \approx \sqrt{2}10^{-k}$$<br /> and<br /> $$p \approx \frac{m_0 v \times 10^{k}}{\sqrt{2}}$$<br /> so<br /> $$p \approx 10^{k-19} \frac{kgm}{s}$$<br /> (This is all of course, very rough).<br /> Now<br /> The mometum of a $$50g$$ bullet trave ling at $2000 \frac{m}{s}$is<br /> roughly $$10^2 \frac{kgm}{s}$$<br /> so if the proton is traveling at $$c \times (1 - 10^{23})$$ then it will have roughly the same amount of momenum as a bullet, and, if you manage to stop a proton traveling that fast, it will probably kill you because the amount of energy involved is pretty large... $E \approx pc \approx 10^{10}J$ or $$10^30 eV$$ (I think that's roughly a ton of TNT).[/itex]

 

[chroot]

NateTG:

That's enormously faster than the original poster's 0.9999c figure. :)

- Warren

 

[Tachyon son]

I indicate 0.9999 c because, as far as I have read, is the speed achieved by a modern accelerator (i.e, tevatron).

I know that in that range a "bit" more speed near c increase massively the mass of the proton. I suppose it will be good to state a definitve speed in this topic to make the correct calculus.

 

[NateTG]

I indicate 0.9999 c because, as far as I have read, is the speed achieved by a modern accelerator (i.e, tevatron).

I know that in that range a "bit" more speed near c increase massively the mass of the proton. I suppose it will be good to state a definitve speed in this topic to make the correct calculus.

If
$$v=(1-10^{-4})c$$
then
$$p\approx 10^{-15} \frac{kgm}{s}$$
and
$$E \approx 10^{-7} J$$

Neither of which are particularly daunting numbers - certainly not enough to cook a human. Of course, as far as I am aware particle accelerators do not, typically, produce single accelerated protons. At 500kW beam power there's still enough heat to cause some serious problems.