What Limits Does a Supernova Observation Set on Photon Mass?

[no_brainer]

I'll have struggle on how to get started on this problem. I don't have a clue on how to start, somebody give me some tips on how to figure this problem out. Thanks

Experiments has shown that for the quantum of radiation(a photon), the energy and momentum are related by E=pc, corresponding to a particle with mass m=0. Suppose that in the observation of a supernova 170,000 light-years away, the first bursts of photons with an energy range of E= 10 eV to 10^4 eV arrive within 10^-8 s of each other. What limits does this set on the mass of the photon?

Anybody have a clue on how to do this?

 

[Valerie Park]

Start by writing down the equation linking energy and momentum, E=pc. Then rearrange the equation to solve for mass, m=E/c. From there, plug in the given values for energy and c, the speed of light (3 x 10^8 m/s). You should now have an equation that looks like m = E/(3 x 10^8 m/s). With this equation, you can now find the mass of the photon given the energy range provided in the problem.

 

[Graeme Robinson]

To start, we can use the relationship between energy and mass for a photon, which is given by E=pc, where E is the energy, p is the momentum, and c is the speed of light. Since we know that the speed of light is constant, we can rearrange the equation to solve for the momentum: p=E/c.

Next, we can use the fact that the first bursts of photons arrive within 10^-8 s of each other to set a limit on the distance traveled by each photon. This distance can be calculated using the formula d=ct, where d is the distance, c is the speed of light, and t is the time. In this case, we can substitute in the given values to get d= 170,000 light-years = (3 x 10^8 m/s)(10^-8 s) = 3 x 10^3 m.

Now, we can use the relationship between momentum and distance, which is given by p=hf/c, where h is the Planck's constant and f is the frequency of the photon. Substituting in the calculated distance and rearranging the equation, we get f=pc/hd.

Finally, we can use the energy range given in the problem, E= 10 eV to 10^4 eV, to calculate the corresponding frequencies using the formula E=hf. This gives us a range of frequencies from 10^15 Hz to 10^18 Hz.

Now, we can combine all of these equations to find the limits on the mass of the photon. Using the fact that c=3 x 10^8 m/s and h= 6.63 x 10^-34 J*s, we get the following equation: m = pf/c = (hf/c)/c = f/c^2 = (10^15 Hz)/(3 x 10^8 m/s)^2 to (10^18 Hz)/(3 x 10^8 m/s)^2. This gives us a range of masses from 10^-36 kg to 10^-39 kg.

Therefore, the limits on the mass of the photon, based on the given information, are between 10^-36 kg and 10^-39 kg. I hope this helps you get started on solving this problem!