Since school and college, I have loved to rearrange formulae. Its actually a very good way to learn about the relations between various quantities (for instance F=ma and p=m/v and m=pv, therefore F=pva, you can see the relationship between force and density, in an abstract sense at least).
But rearranging E=mc^2 to express c makes it more confusing. (and I am not even going to try to replace m with F/a since that gives me a headache.
If c is fixed (or in other words, the laws of physics are accurate) then it should have no relationship to E or m.
What am I missing here? (Besides the obvious, having a proper physics education).
Hyper-grossly simplifying stuff, in your equation, c is a constant. It has a known, fixed value, and it’s written as c because it’s a lot shorter than 3.00x10^8 m/s, which you surely know of course.
Reframing this to an equation you’re surely more familiar with, this is akin to the equation relating distance to uniform acceleration, (d=1/2at^2) and rearranging it all to solve for 1/2 (1/2=d/at^2.)
The 1/2 is a byproduct of the calculus from integrating d=vt; but it would also hold up to scientific scrutiny with relatively little effort, if you could arrange for something to experience uniform acceleration over a known time. (Gravity in a vacuum will serve nicely.). At least these are values that are easily measured.
You could scientifically ascertain the value of c if you could arrange for a spontaneous matter/antimatter reaction of known masses and could measure the resulting energy output. That’s how you’d get known E and m values and could drop them into the equation to see what you get.
This is getting longer than I meant to write. It would be worth your time, because it’s genuinely interesting, to try and find a math-light explanation of the purpose of the equation.
To put it another way, c doesn’t have a relationship to E or m. E (the rest energy of an object) and m (the mass of that object) have a fixed relationship of c2.
You can do the same thing with, say, pi. If A = πr2 (area of a circle), you can “solve” that for pi. You’ll get π = A/r2. Does that mean the value for pi depends on the area and radius of a circle? No. It just means they have a fixed relationship. If you change r, then A has to change.
Same with the rest energy and mass. They have a fixed relationship. A certain amount of mass always has the a certain amount of rest energy. If you change m, E must also change by a factor of c2.
Let’s back up a bit. First of all, you can always check your own work by expressing physical quantities in terms of base units, in this case mass, length, and time. So mass has the unit M , c is a speed therefore has units LT−1, and energy has units ML2T−2.
So, c is not a fixed dimensionless constant: it depends on the units used to measure mass and time. (And so do the other quantities in your equation.) That is why it has a relation to E and to m; any change made to the units used to measure them will change the value of c as well.
That’s probably the simplest way of looking at it. The OP’s restatement of E=mc2 might seem odd because m and E are arbitrary quantities; indeed they are, but they are not independently variable. They have a fixed relationship and fixing a value for one automatically fixes the value of the other. The original form of the equation says that if you know an object’s mass, you can derive its total energy equivalent.
Conversely, quantum particles in particle accelerators are conventionally described in terms of electron volts, nominally a unit of energy, but which also defines their mass due to the above relationship – one eV is equal to 1.782662×10−36 kg.
You could also think about it like a right triangle whose hypotenuse is 4. The other two sides can vary, but only as long as
2 = sqrt(a^2+b^2)
which of course if course is just rewriting the pythagorean formula.
Just because a way of writing an equation is algebraically true doesn’t necessarily make it interesting or useful. You could, I suppose, measure the value of c by measuring the mass and energy of some particle, in which case that form of the equation would be useful, but there are other methods for measuring c that are both easier and more precise.
The technique is called dimensional analysis - what are the basic units of the equation, and do they balance?. M is mass, e is energy. Energy (aka work) is force times distance. force is mass times acceleration. Acceleration is change of velocity with time (i.e. meters/sec^2). Velocity is distance over time.
c= sqrt( (distance x (mass x (distance/(time x time))) / mass)
c= sqrt ( mass x distance x distance /(time x time)/mass) and the mass units cancel out.
so c is in units (distance / time) which, considering it’s the speed of light, is obviously true.
It’s a constant but depends on what units you use to express it. m/sec? mph? miles/sec?
i.e. the freezing point of water is a constant but you can either call it 0°C or 32°F etc.
There are a few basic units - mass, distance, time, temperature, etc. Most other measurements are ratios of these.
Since the main question has been answered: If by p you mean momentum, then p = mv. I think the most interesting way to combine that with F = ma is to write a = dv/dt so that F = m dv/dt and, assuming m is constant, F = d(mv)/dt = dp/dt. In a sense, this is more fundamental than F = ma: force is (produces) change in momentum.
Let me try another way of answering the OP:
Well, yeah, sure.
And in the unlikely event that you lose access to the value of c, but have an example of E and m in a conversion, why, you’re all set.
In Newtonian dynamics, conservation of energy, momentum and mass are seperate conservation laws; in special relativity these are replaced by a single consevration law - the conservation of 4-momentum. Relativsitc conservation of momentum is the conservation of the spatial components of the 4-momentum vecotr. Relativisitic conservation of energy is the conservation of the temporal component of 4-momentum. Conservation of mass is replaced by the conservation of the norm (the ‘length’) of 4-momentum, which is the same as the conservation of ‘rest energy’. The rest energy of a system can be entirely locked in mass, and c^2 is just a conversion factor to get convert between units of energy and mass.
It isn’t possible to derive E =mc^2 from Newtonian formulae such as F=ma. The basic argument to get E=mc^2 is that it is necessary to have this relationship in reltivistic dynamics in order to recover Newtonian dynamics in the correct limit. That is why Einstein’s ‘proof’ (which us Maxwellian electrodynamics) of E=mc^2 is more a heuristic explanation than a proof.
The SI unit of energy and work is the Joule (J), which is, in high school physics, usually introduced as Newton multiplied by metre - if you pull something with a force of 1 Newton, and you do that over the length of a metre, then you have performed work of 1 J (or spent eneergy of that amount). But the Joule can also be looked at as a unit defined as kg*m² / s². That definition gives you a unit of mass (kg), multiplied by the square of a unit of velocity (m/s), which confirms that the connection between mass, velocity and energy expressed in the formula really is right. You have the same connection in the standard formula for the kinetic energy of a given object moving at a given velocity, E = 0.5m * v² (here, m stands for mass, not for the unit of distance).
Coincidentally, the Newton is defined as kg * m/s² (it takes 1 N of force to give a mass of 1 kg an acceleration of 1 m/s²); so if you multiply this by a metre, you get the same unit of energy, kg * m² / s². Which closes the circle to the earlier definition of the J, exerting a force of 1 N over a distance of 1 m.
Lastly, as far as electricity is concerned, work of 1 J is also done if an electric current of 1 A (Ampere) at a voltage of 1 V (Volt) runs for 1 second. It’s nice to see how all these units fit together.
Nothing coincidental about that: The SI units were designed to work together in that way. That’s one of the three big benefits of SI over the American system.
The shortest possible answer is that E and m have a fixed relationship to each other which is the whole point of this equation.
In a very real sense, c here is just a unit conversion factor. (See the many, often bumpy, past threads on this.) So while it is mathematically fine to rearrange the equation to isolation c, it’s sort of a weird thing to ever want to do. A analogy would be taking:
(height in centimeters) = 2.54 * (height in inches)
and writing it as:
2.54 = (height in centimeters) / (height in inches) .
It’s always true, but it’s a weird way of expressing “2.54”.
I remember learning how to attack a physics problem by setting all the physical constants in the equation to “1” and then at the end multiplying or dividing by these constants until the dimensional units of the answer were right.
Yes, it actually does work.
The way we learned it, more generally you do a little algebra to divide your parameters into parameters with independent dimensions and parameters whose dimensions can be obtained as products of powers (i.e., multiplying and dividing) of the first set. For example, suppose you have fluid flowing through a pipe, and you want the pressure drop per unit length as a function of the mean fluid velocity, the diameter of the pipe, the fluid density, and the fluid viscosity: dp/dx = f(U, D, ρ, μ) . In this case the dimensions of the first three parameters are independent, while the units of viscosity are the same as density times velocity times length. Now you can multiply and divide to rewrite the relation in terms of dimensionless parameters: set Π = (dp/dx)/U2D−1ρ and Π₁ = μ/UDρ, so that Π = U−2Dρ−1 f(U, D, ρ, UDρ Π₁) = Φ(U, D, ρ, Π₁), where Φ is some function. The upshot of this dimensional analysis is that Φ does not depend on U, D, or ρ, so we finally obtain Π = Φ(Π₁), that is, dp/dx = U2D−1ρ Φ(μ/UDρ). So even though in this case we cannot reduce to a constant (zero variables), we see that the pressure drop can be expressed in terms of a function of only one variable, the Reynolds number.
That’s your coherent unit system for you.
And it’s a point missed by people who argue that feet are a more natural unit than meters, that pounds make more sense than other units, that Fahrenheit temperatures are easier to judge. They aren’t. But, worse, they create so much more work for technical people all over the world, and so many opportunities for little mistakes that make the whole shebang just wrong wrong wrong. There are over 200 different possible units for heat transfer coefficient (which I work with a lot) in the English system and its various associates, and just one in the SI.
I read that, and immediately remember the Mars probe that crashed because of a missed english/metric conversion. Good times.
Similarly for a plane that ran out of fuel because of confusion between kilos and pounds, and fortunately did not crash.