The relationship between mc^2 and mc^2 x 1/2

[thetexan]

TL;DR

Comparison of two formulas

Kinetic energy = 1/2 m V^2

I was thinking about this and thought another formula…

E = mc^2

These look very similar except for the multiplication by 1/2.

Let’s say you take a kilogram ball of uranium and accelerate it to the speed of light. I know, I know. You can’t. But let’s say you did. We could calculate its kinetic energy with E=mc^2 then dividing by half.

So it seems that the difference between the kinetic energy of the ball of uranium traveling at the speed of light is 1/2 of the ball’s mass totally converted to energy as per E=mc^2.

Is this correct?

Tex

 

[Ibix]

Is this correct?

You answered your own question:

I know, I know. You can’t.

If you consider something that is impossible according to the laws of phyics, why would you expect a coherent answer about what happens from those laws? It's like asking "if a banana were a ham sandwich what would it taste like - a banana or a ham sandwich?" The premise of the question is nonsense and any answer I can state, like "it's still a banana even though it's a ham sandwich so it would taste like a banana", would be just empty words. You started from inconsistent premises so your output is nonsense - using maths just makes it less intuitively obvious that the output is nonsense.

Also, the correct expression for kinetic energy is $K=(\gamma-1)mc^2$ where $\gamma=\frac 1{\sqrt{1-v^2/c^2}}$. In the case where $v\ll c$ you can Taylor expand $\gamma$ to get $\gamma\approx 1+v^2/2c^2$, which is where the Newtonian $K=\frac 12mv^2$ comes from. It is not valid at high speeds, and you need to use the correct expression. However, the correct expression is undefined at $v=c$ because it is not possible for a massive object to travel at that speed.

 

[kuruman]

@Ibix prempted me by saying essentially what I was going to say so here are some numerical calculations as an illustration.

If you accelerate the uranium to $v=0.9999c$, its kinetic energy will be $K=70mc^2$
if you accelerate it to $v=0.99999c$, its kinetic energy will be $K=223 mc^2.$

Clearly, if you add more nines past the decimal to get closer to the speed of the light, you will not get to $\frac{1}{2}mc^2$ because you are already way past that point.

BTW, 1 kg of mass from your recycling bin works just as well and is more readily available than uranium.

 

[thetexan]

I understand that it’s an absurd hypothetical. My curiosity is, I admit, due to my ignorance. But the equation which I thought was correct…ke=1/2mV^2…seemed to accept any value for V, whether 100mph or 400000000 miles per second (well over c), without regard of the impossibility of exceeding c. So if you use the max (c) for V then it would seem to give the kinetic energy for that mass at c. My point (albeit apparently flawed) was that the kinetic energy due to velocity would be exactly 1/2 of that if the mass were to be totally converted to energy using e=mc^2. That seemed very unusual to me and oddly coincidental.

Based on the reply the formula does not work for very high speeds. At approximately what velocity does this breakdown in formula usability occur?

Tex

 

[kuruman]

Based on the reply the formula does not work for very high speeds. At approximately what velocity does this breakdown in formula usability occur?

In depends on what you mean by breakdown. What should the discrepancy between the two be before one declares "breakdown"? 1%, 5%, 20%? See discussion about relativistic energy and plot here.

 

[Ibix]

My point (albeit apparently flawed) was that the kinetic energy due to velocity would be exactly 1/2 of that if the mass were to be totally converted to energy using e=mc^2. That seemed very unusual to me and oddly coincidental.

For any energy that is a mass times something, the something needs units of velocity squared. So if you plug in the velocity $c$ into two such formulae the answers necessarily have a dimensionless ratio that's usually fairly simple.

Based on the reply the formula does not work for very high speeds. At approximately what velocity does this breakdown in formula usability occur?

Depends on how precise you want to be, as @kuruman says. As I said above, you get the Newtonian expression by keeping the first two terms in the Taylor expansion of $\gamma$ and saying that the rest of the terms are too small to worry about. If you keep the next term too then you get that $$\gamma\approx 1+\frac{v^2}{2c^2}-\frac{3v^4}{4c^4}$$Plug that approximation in to the $K=(\gamma-1)mc^2$ relativistic kinetic energy equation and you get$$K\approx \frac 12mv^2-\frac 34m\frac{v^4}{c^2}$$which is a slightly better approximation than the Newtonian one. The difference is the last term, which is on the order of $v^2/c^2$ smaller than the first one. When the velocity is high enough that $3mv^4/4c^2$ is about one division on your energy measuring device's scale, then you probably need to use the relativistic formula. If you trust the device to three significant figures the threshold will be somewhere in single digit percentages of light speed. But if your kit is better or worse than that it might be higher or lower.

 

[jbriggs444]

Based on the reply the formula does not work for very high speeds. At approximately what velocity does this breakdown in formula usability occur?

One experiment that clearly shows the discrepancy is the Bertozzi experiment using high speed electrons.

It is worth a watch.

 

[kuruman]

It is worth a watch.

It really is. Thanks for posting. They don't make them as thorough as this. There is something nostalgic about the educational value of real experimentation that successfully makes its point with analog data collection and analysis. It's all in the design. I still remember the concluding words of a talk by J.A. Wheeler that I heard many many years ago, "Do what you can, with what you have, where you are." He said it twice to make sure it stayed with the audience.

 

[Ibix]

It is an excellent physics video. Also, bring back suit and tie in the lab!

 

[gmax137]

This is an excellent video, I hope the OP takes the time to watch.

PS, if this vid doesn't inspire you to study physics, I don't know what will.

 

[Mark44]

Let’s say you take a kilogram ball of uranium and accelerate it to the speed of light. I know, I know. You can’t. But let’s say you did. We could calculate its kinetic energy with E=mc^2 then dividing by half.

Don't you mean "then multiplying by half"? Dividing by 1/2 is the same as multiplying by 2.

 

[weirdoguy]

PS, if this vid doesn't inspire you to study physics, I don't know what will.

Well, I'm one of those (few?) physicists that do not get excited by that kind of things at all. But I do get "aroused" by the idea of using affine bundles in formulating Newtonian mechanics And I was that way from the very moment I started falling in love with physics, back in high school. Each to their own, I guess.

Dividing by 1/2 is the same as multiplying by 2.

I don't know how it is in other languages, but in polish, in common language, "divide by half" and "divide by two" means exactly the same

 

[jbriggs444]

I don't know how it is in other languages, but in polish, in common language, "divide by half" and "divide by two" means exactly the same

In English, "divide by half" is not a phrase that one often sees used. "Divide by two" or "Divide in half" would be typical for people whose goal is effective communication.

Just the other day, I saw "divide by half" used by someone whose goal was not effective communication. I think it was a youtube short. He wanted to look clever by pointing out the unexpected meaning.

 

[berkeman]

I don't know how it is in other languages, but in polish, in common language, "divide by half" and "divide by two" means exactly the same

Shouldn't that be "reverse polish" notation?

 

[Klystron]

One experiment that clearly shows the discrepancy is the Bertozzi experiment using high speed electrons.

It is worth a watch.

Excellent painstaking experiment leading inexorably to the conclusion that speed of light is the limiting velocity as more energy is pumped into the electron bunches. I particularly like verifying energy input with thermal data from the aluminum target.

 

[sophiecentaur]

I don't know how it is in other languages, but in polish, in common language, "divide by half" and "divide by two" means exactly the same

So one should avoid using "common language" in important situations. There's a similar sloppy way of talking when people say "ten times less" which can result in misunderstanding. My dear old Grandad used to ask me 'what's twice the half of one and a half?"

I read, recently, that a burger chain tried unsuccessfully to get people interested in One third pounders but they though a quarter pounder was actually more.

That's the sort of maths that would suit Rishi Sunac's idea of extending maths education to older kids.

 

[jbriggs444]

a burger chain tried unsuccessfully to get people interested in One third pounders

Google finds a reference to A&W:

The firm eventually conducted a focus group to discover the truth: participants were concerned about the price of the burger. "Why should we pay the same amount for a third of a pound of meat as we do for a quarter-pound of meat?" they asked.

It turns out the majority of participants incorrectly believed one-third of a pound was actually smaller than a quarter of a pound.