# This is probably just a coincidence, but

1. Jan 13, 2006

### Trepidation

$$EK = mv^2$$
energy kinetic = mass times velocity squared

So... Isn't one way to interpret time dilation that everything moves through time and space with a sum velocity of c (somehow), and that as velocity in space increases velocity in time therefore decreases? I've read this in several places, anyway.

If this is the case, then the permanent space-time velocity of all objects is c... Which means that their kinetic energy would be:

$$EK = mv^2$$

$$EK = mc^2$$

So er... $$E=mc^2$$

2. Jan 14, 2006

### Mortimer

The equation $E=mc^2$ only applies when $v=0$. The general equation for any $v$ is
$$E=\sqrt{(mc^2)^2+c^2p^2}$$
where $p$ is momentum.
So indeed just a coincidence.

3. Jan 14, 2006

### ZapperZ

Staff Emeritus
1. Since when is kinetic energy equal to mv^2? What happened to the 1/2?

2. What exactly are "velocity in space" and "velocity in time"?

3. What is "space-time velocity"?

4. This statement is puzzling: "..... Isn't one way to interpret time dilation that everything moves through time and space with a sum velocity of c (somehow),... " Everything does NOT move with a "sum velocity of c (somehow)".

Zz.

4. Jan 14, 2006

### Mortimer

Of course. I completeley overlooked this most obvious argument!

5. Jan 15, 2006

### franznietzsche

No, its $$E_K = \frac{1}{2} m v^2$$

$$E=mc^2$$ only applies to a an object when $$v= 0$$ as another pointed out. It has nothing to do with kinetic energy.

6. Jan 15, 2006

### masudr

The length of the velocity four-vector of any particle is 1 in geometric units, which corresponds to c in SI units. A particle in it's rest frame is moving through time at a second a second. To compare lengths with times, we multiply by c, to find that it is moving c metres a second.

7. Jan 15, 2006

### ZapperZ

Staff Emeritus
But really, do you honestly think the OP knew about this and that this is what he/she is describing by making that erroneous statement? I highly doubt it.

Zz.

8. Jan 15, 2006

### masudr

I believe one of Greene's recent popular science texts (with some grandiose title) expresses this idea (and I assume this is where the OP got the idea from); whether or not the OP fully understood it or not is, of course, another matter.

9. Jan 15, 2006

### ZapperZ

Staff Emeritus
I'm not so sure... Greene cannot make a silly mistake of equating KE with mv^2 and missing out that 1/2, which is the OP starting premise.

Zz.

10. Jan 15, 2006

### Trepidation

I comprehend the idea, but the important thing is that I apparently don't comprehend simple Newtonian physics. Forgive me for the incorrect formula and the idiotic post... I just saw something that sort-of corresponded and then decided to post it, and consequently wound up with a bunch of gibberish.

Again, sorry. I don't know what I thought when I was writing EK=mv^2. Thanks for replying to this, anyway.

11. Jan 15, 2006

### masudr

Sorry, I meant the notion of having a velocity of c at all times, as opposed to the definition of non-relativistic classical kinetic energy.