What is a joule, when you calculate relativistic energy?

In summary, the energy of a photon can be calculated using the formula E=pc, where E is the energy and p is the momentum. Since photons have zero rest mass, their mass when moving at the speed of light is also zero. However, their energy and momentum can still be measured through interactions. The terms PJ and MJ are most likely referring to petajoules and megajoules, respectively, but without context, they can be ambiguous.
  • #1
Dgonzo15
14
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Energy is E=γmc^2, but when I calculate this, will my result be in joules? I am unsure what the units are when I calculate it, and I keep hearing people saying joules.
Also, what is PJ and MJ?
 
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  • #2
Dgonzo15 said:
Energy is E=γmc^2, but when I calculate this, will my result be in joules? I am unsure what the units are when I calculate it, and I keep hearing people saying joules.
Also, what is PJ and MJ?

Relativity doesn't change the units of anything. If you use kg, meters, seconds you get joules for energy. Most commonly, I would guess PJ is petajoule and MJ megajoule, but these are ambiguous out of context: MJ could be millijoule or microjoule; PJ could be picojoule or the founder of groklaw.
 
  • #3
So how would I calculate the energy of a photon? Since a photon travels at c, gamma (in E=γmc^2) would be undefined, so it doesn't make any sense to calculate the energy of a photon.
Also, are photons considered to have mass, because mass is also a factor in E=γmc^2.
 
  • #4
Dgonzo15 said:
So how would I calculate the energy of a photon? Since a photon travels at c, gamma (in E=γmc^2) would be undefined, so it doesn't make any sense to calculate the energy of a photon.

Not if you use an incorrect formula for a photon, no. :wink:

The correct formula, valid for any object, timelike (nonzero rest mass, moves slower than light) or lightlike (zero rest mass, moves at speed of light) is:

[tex]E^2 = p^2 c^2 + m^2 c^4[/tex]

where ##E## is the energy, ##p## is the momentum, and ##m## is the rest mass. A photon has ##m = 0##, so the formula reduces to ##E = pc##.
 
  • #5
Dgonzo15 said:
So how would I calculate the energy of a photon? Since a photon travels at c, gamma (in E=γmc^2) would be undefined, so it doesn't make any sense to calculate the energy of a photon.
Also, are photons considered to have mass, because mass is also a factor in E=γmc^2.

You need to use the more general formula: E^2 = (mc^2)^2 + p^2 c^2 [p is momentum]

For massless particles you get E=pc. For a photon, E = h[itex]\nu[/itex], p=E/c.

Note, if you plug in p = [itex]\gamma[/itex]mv, you will get the special formula for massive particles.
 
  • #6
Dgonzo15 said:
Also, are photons considered to have mass, because mass is also a factor in E=γmc^2.

Photons have zero rest mass. Some people use the term "relativistic mass", but that's really just another term for "energy"; a photon certainly has energy, so it does have relativistic mass, but as in my previous post, you can't define its relativistic mass/energy using ##\gamma##, so you have to use the more general formula I posted.
 
  • #7
OK, so the formula E^2=P^2... reduces to E=pc for photons since photons have zero REST mass. Since they have zero REST mass, what is their mass when they are moving at c? You would need to know their mass in order to calculate p so you can calculate E when they are moving, right?
 
  • #8
Dgonzo15 said:
OK, so the formula E^2=P^2... reduces to E=pc for photons since photons have zero REST mass. Since they have zero REST mass, what is their mass when they are moving at c? You would need to know their mass in order to calculate p so you can calculate E when they are moving, right?

No, you directly measure momentum via interaction (using conservation of momentum). There is not mass distinguishable from energy (which is also directly measurable). For light, relevant formulas use only E and p not mass.

Note, this does not imply photons are not affected by gravity. In GR energy is a source of gravity and is affected by gravity.

Really, what distinguishes mass from energy is the existence of a frame where p=0, and KE=0. Without that, there is no meaningful way to distinguish mass from energy.
 

1. What is a joule and how is it related to relativistic energy?

A joule is a unit of energy, named after the English physicist James Prescott Joule. It is equal to the amount of work done by a force of one newton acting through a distance of one meter. In the context of relativistic energy, a joule is the unit used to measure the amount of energy an object has due to its motion relative to an observer.

2. How is relativistic energy calculated using joules?

Relativistic energy is calculated using the famous equation E=mc^2, where E represents the energy, m represents the mass of the object, and c represents the speed of light. The unit of energy, joules, is used to express the value of E in this equation.

3. Can joules be used to measure both kinetic energy and potential energy in relativistic contexts?

Yes, joules can be used to measure both kinetic energy and potential energy in relativistic contexts. Kinetic energy is the energy an object possesses due to its motion, and potential energy is the energy an object has due to its position in a force field. Both of these types of energy can be measured in joules in the context of relativistic energy calculations.

4. How does relativistic energy differ from classical energy calculations using joules?

Relativistic energy calculations differ from classical energy calculations in that they take into account the effects of special relativity, such as time dilation and length contraction. In classical energy calculations, these effects are not considered and the equations used are simpler. However, in extreme cases where objects are moving at very high speeds, relativistic energy calculations are necessary for accurate results.

5. Are joules the only unit used to measure relativistic energy?

No, joules are not the only unit used to measure relativistic energy. Another commonly used unit is electron volts (eV), which is equal to the amount of energy gained by an electron when it is accelerated through a potential difference of one volt. This unit is often used in particle physics and nuclear physics, where energies involved are very high.

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