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# How do you solve for mass (without assuming it is zero) E^2 = (pc)^2 +(mc^2)^2

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How do you solve for mass (without assuming it is zero) E^2 = (pc)^2 +(mc^2)^2 If the target wavelength of light is 685 nanometers? E= energy of a photon p= momentum of a photon c= speed of light m=mass I just don't know what to do with all the measurements when multiplying and dividing or where to even begin with what to divide at first.

Nov 7, 2020

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What "mass" are you trying to calculate? Don't confuse the two "E's", energy equations:

1 - E =hf, where h=Planck's Constant=6.626 x 10^-34 Js. f=Frequency.
f =c/λ, where c=speed of light, λ =Wavelength of the photon.

E =hc/λ  [This is the equation that neasures the energy of a photon].

2 - E =MC^2, where M =Mass, C=speed of light. [This is the famous Einstein's equation, equating the mass(of an object) to its energy].

Nov 7, 2020
#2
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Well as a general rule of thumb i've come to recognize is that to get zero you either deduct the exact sum of the parts or multiply by zero. By my calculations there is no deduction that is happening equal to the sum of the parts and you are certainly not multiplying by zero, so light must have mass. Short answer is i'm trying to figure out the mass of light with a wavelength of 685

Guest Nov 7, 2020
#3
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But a "photon of light" DOES NOT have any mass!! That is the reason light travel at its well-known speed, which is the "maximum" speed possible. A photon may have "wavelength" or "frequency" and , of course, "energy", but no "mass".

Guest Nov 7, 2020
edited by Guest  Nov 7, 2020
#4
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Then perhaps you can demonstrate the multiplication of zero or the deduction of the total sum of it's parts using the formula provided and back up this claim.

Guest Nov 7, 2020
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Like so

Nov 7, 2020
#6
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Can you explain how you switched the sides of the m, E, and most importantly how you came upon a minus sign for that nifty trick?

Guest Nov 7, 2020
#7
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Honestly, could you just show the entire process, I'm still confused as to how you divided by c^2 without dividing the other side as well. Plus I just don't get where the minus comes from.

Guest Nov 7, 2020
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1.  Subtract (p.c)2  from both sides. to get  E2 - (p.c)2 = (m.c2)2

2.  Take the square root of both sides to get $$\sqrt{E^2-(p.c)^2}=m.c^2$$

3. Divide both sides by c2 to get $$\frac{\sqrt{E^2-(p.c)^2)}}{c^2}=m$$

4. Substitute for E and p using the expressions given.

Alan  Nov 7, 2020
#9
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Awesome thank you so much. This information is hard to find. I guess not too many people are interested in it, because it's a fact that light is always moving and hence doesn't have a rest mass.

Guest Nov 7, 2020