# Using Relativisitic Units (where c = 1)

• Chetlin
In summary: So there is no problem with having a numerical factor when doing a conversion.In summary, there are different ways to express energy in special relativity, with the most common being E = \sqrt{p^2 + m^2}. This often involves setting the value of c to 1 and expressing momentum and mass in electron-volts. However, other units and values for c can also be used, as long as the conversion is remembered. It is important to remember the chosen conversion rule in order to get back to the original equation.
Chetlin

## Homework Statement

It's not really an assigned problem, but it was part of one, and it's a concept that I'm not quite getting.

## Homework Equations

In special relativity, energy can be expressed using the equation $E^2 = (pc)^2 + (mc^2)^2$, or $E = \sqrt{p^2c^2 + m^2c^4}$. Apparently, according to my textbook and professor and other things I've read, people often set the value of c to 1, leaving the energy equation as $E = \sqrt{p^2 + m^2}$. And, often, p and m are expressed in electron-volts, even though the "true" units would be eV/c for p and eV/c2 for m.

## The Attempt at a Solution

In a question I had, I was given the mass of something in electron-volts, and eventually found its momentum, also in electron volts using 4-momentum techniques. It asked for the energy. Let's be general and say that the mass was m eV and the momentum was p eV (the variables have the units of electron-volts). Are you able to just say that the energy is $\sqrt{p^2 + m^2}\:\:e\text{V}$? I did not feel comfortable doing that and at the end of class today I tried asking my professor if that was actually what we were supposed to do, but it was rushed since it was the end of class and everyone was moving and the only thing I think he understood was that I wasn't comfortable with relativistic units (and it's true, I'm really not, and that's what he told me). As the solution to the problem I gave the very inelegant-looking answer $\sqrt{(p\:\:e\text{V}\!/\!c)^2c^2 + (m\:\:e\text{V}\!/\!c^2)^2c^4}$ which I notice does seem to equal the same thing as the other answer, with all the c instances cancelling. But ... it just doesn't "seem" right to say that mass and momentum are in energy units and then add them to get energy units, since mass and momentum don't have the same units really and can't add. Is this the actual "right" answer or is the first way just as fine?

both ways are fine. When you use something like relativistic units, you just need to remember what you used to turn the units from the usual SI units into the natural units. In this case, you just used one 'conversion' which is c=1. So if you have any equation in the natural units, then you can get back to the SI units by getting the dimensions correct by inserting 'c' into the places where it needs to be.

In other words, there is a unique way to convert between the two systems of units. So therefore, you can think of any equation, written in the two different set of units as being 'essentially the same equation'. However, if you forgot that you used c=1. (for example, you might have used c=2 but can't remember either way), then you cannot get back to your original equation. So it is important to remember what rule you used to do the conversion in the first place.

1 person
Thanks! I understand this a lot more now.

That's something I considered too, setting c = 2 or something. It seems that if you do that, you can't express p and m with the same units now. The units for p become eV/2 and those for m become eV/4. The only way I can see to work with these kinds of units is to basically multiply things again to get it to the same form we had when c = 1. I know it's not "normal" to set c equal to anything else, but I know it's still something that can be done. However, it's tough to see how doing that wouldn't make the calculation messier than just leaving every c in from the beginning.

yeah, In this case, c=1 is definitely the nicest option, but as you say, we are still free to make whatever choice we want. As an example of when the conversion involves a numerical factor, there are Gauss-cgs units, which have a factor of ##\sqrt{4\pi \epsilon_0}## to convert the electric field from Gaussian-cgs units to SI units. So, in this case, they have a numerical factor of ##\sqrt{4\pi}## which is just used to make the equations a bit nicer.

I can understand your confusion with using relativistic units and the equation E = \sqrt{p^2 + m^2}. It may seem counterintuitive to add quantities with different units to get an energy, but in the context of special relativity, this approach is valid and commonly used.

The reason for this is that in special relativity, mass and energy are seen as interchangeable and can be expressed in the same units. This is due to Einstein's famous equation E = mc^2, which relates mass and energy. In relativistic units, the speed of light (c) is set to 1, so mass and energy can be expressed in the same units (e.g. electron-volts).

In your example, the mass and momentum are both given in electron-volts, so they can be added together to get the energy in electron-volts. This is because in relativistic units, momentum (p) is defined as p = mv, where v is the velocity, and since c = 1, p and m have the same units.

So, to answer your question, yes, you can say that the energy is \sqrt{p^2 + m^2}\:\:e\text{V}. This is a valid approach in relativistic units. However, if you prefer, you can also use the full expression with c to make it more clear that you are adding quantities with different units. Ultimately, both approaches will give you the same answer.

I hope this helps clarify the use of relativistic units and the calculation of energy in special relativity. Keep practicing and asking questions, and you will become more comfortable with this concept.

## 1. What are relativistic units?

Relativistic units are a system of measurement in which the speed of light, c, is set to equal 1. This means that all other units of measurement, such as time and distance, are also scaled accordingly.

## 2. Why use relativistic units?

Relativistic units are commonly used in physics and other fields of science because they simplify mathematical equations and make them easier to work with. They also allow for a better understanding of the relationship between different physical quantities.

## 3. How do you convert between relativistic units and traditional units?

To convert between relativistic units and traditional units, you can use the conversion factor c = 1. For example, to convert a distance of 10 meters to relativistic units, you would divide by the speed of light, c, which gives you a value of 10/299,792,458 = 3.34 x 10^-8 relativistic units.

## 4. What are some common examples of relativistic units?

Some common examples of relativistic units include the meter (m) for distance, the second (s) for time, and the kilogram (kg) for mass. However, any unit of measurement can be converted to relativistic units by dividing it by the speed of light, c.

## 5. What are the advantages of using relativistic units?

Using relativistic units allows for a more intuitive understanding of the relationship between physical quantities, simplifies mathematical equations, and is often used in theories and models of physics, such as Einstein's theory of relativity. It also eliminates the need for conversion factors when working with equations involving the speed of light, c.

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