What Is the Laboratory Frame Energy of a Proton?

[cartonn30gel]

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

 

[blochwave]

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

 

[Moetasim]

Homework Statement

Homework Equations

The Attempt at a Solution

Sorry, I am not able to solve this problem. It is beyond my level of expertise in relativity. I think you should post the problem in the Advanced Physics section to get a better response.

The question is asking about the energy of the proton in the laboratory frame, not its speed. To find the energy in the laboratory frame, you need to use the relativistic energy equation: E^2 = (pc)^2 + (mc^2)^2

In this case, the energy of the proton in the laboratory frame is given by:

E^2 = (pc)^2 + (mc^2)^2
= [(5MeV/c)^2 + (m_proton*c^2)^2]
= (25MeV^2/c^2) + (m_proton*c^2)^2

You can use the fact that the energy of the photon in the laboratory frame is 35eV to solve for the mass of the proton.