What Is the Laboratory Frame Energy of a Proton?

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The discussion revolves around calculating the laboratory frame energy of a proton in relation to a photon, with the photon having an energy of 35 eV in the lab frame and 5 MeV in the proton's frame. The relativistic Doppler effect is highlighted as a key factor, indicating that the proton and photon are approaching each other at nearly the speed of light. The calculations suggest that the proton's speed is just under the speed of light, leading to an enormous relativistic factor (gamma), which implies that the proton's energy would also be extraordinarily high. The conversation emphasizes that while the calculations suggest an infinite energy scenario if the proton reached light speed, objects with mass cannot attain that speed due to the requirement of infinite energy. Ultimately, the analysis confirms the proton is moving at a velocity very close to the speed of light, resulting in significant energy implications.
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how fast is that proton??

SORRY; I HAD TO MOVE THIS TO THE ADVANCED SECTION

Homework Statement



We are in a laboratory and there is a photon and an ultra-relativistic proton moving. The energy of the photon in the frame of reference of the laboratory is 35eV. The energy of the same photon in the frame of reference of the proton is 5MeV. What is the energy of the proton in the frame of reference of the laboratory?

Homework Equations



energy = h*f
Relativistic doppler effect
Lorenz velocity transformations
relativistic energy

The Attempt at a Solution



using energy = h*f

first frequency (f1) is 8.48*10^(15) Hz
second frequency (f2) is 1.21*10^(21) Hz

change of frequency is I guess because of Doppler effect
because f2 > f1, the proton and the photon must be approaching each other. But no matter what, the proton will see the photon approaching with a speed of c. I am not sure if it is valid to use the Doppler formula here. If I use, I get,
B=v/c=0,99999 (the relative speed of the photon and the proton)

then using Lorenz velocity transformation in x direction I get the speed of the proton to be -c. I thought this could be an indication of the proton's actually to be moving away from the photon (that means both moving in the same direction)

But then I cannot use Energy= (gama)*m*c^(2) because (gama) will be infinite.

I'd appreciate any help.
thanks
 
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I believe you're right to use the doppler effect, but I didn't follow your application

The relativistic doppler shift is frequency observed = sqrt[(1-v/c)/(1+v/c)]*frequency emitted from the source

The problem is V is going to be humongous and yes, negative

f2/f1 is like somethingx10^35, so you get (10^35-1)=-(10^35+1)v/c

Divide both sides by 10^35+1 and you get -v/c=a number just so very much almost 1 but just not quite, like .9999 out to 30+ decimal places

Now when you use the equation like I did, you're assuming positive V means they're moving AWAY from each other, so the negative means the opposite of what you concluded, they're rushing towards each other, and the proton is basically going an incredibly small amount less than c. So yes gamma is going to be gigantic to the point of ludicrous and the energy of the proton is going to be gigantic to the point of unreasonable. But I think that's the idea

If it were ACTUALLY going the speed of light, which you rounded off too, then yes it would have infinite energy.

Which is the point, objects with mass don't get to go the speed of light because we can't impart infinite energy to them. So I think you did the problem right, me following or not, but reached the wrong conclusion
 
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