# Revolutionary physics

1. Aug 26, 2006

### Loren Booda

In your opinion, what physics of the past 100 years most closely approaches the elegance, simplicity and appeal of Einstein's equation E=mc2?

2. Aug 26, 2006

### desA

3. Aug 26, 2006

### 3trQN

I dont know any physics from the last 100 years :s

4. Aug 26, 2006

Staff Emeritus
$$\Delta x \Delta p = ih$$

5. Aug 26, 2006

### K.J.Healey

I agree. This is much more revolutionary than simple relativity.

And to be technical, Einstein first publised his papers in 1905 on SR, which would make E=mc^2 out of the 100 year range given I believe. :tongue2:

6. Aug 26, 2006

### Schrodinger's Dog

I am going to show my supreme ignorance of physics by asking what this means?

The change in x and p = complex number times the plank constant? Wild guesses here? I'm serious I have no idea what this relates too sorry.

For us physics numptys could you let us in on this breakthrough?

EDIT: I cant believe I didn't conjugate the word numpty properly, I apologise to Scotsmen everywhere!

Last edited: Aug 26, 2006
7. Aug 26, 2006

### Loren Booda

The imaninary number i should be left out of that Heisenberg uncertainty principle, but included in such as the Q. M. operator for momentum.

What do you all think of ther contention that relativity is more fundamental than quantum mechanics?

8. Aug 26, 2006

Staff Emeritus

Yes the imaginary i comes in in quantizing the momentum to an operator. Ignore it for the present purposes. The change( or range of change ~ standard deviation of distribution of possible values) for the position x, multiplied by the same thing for the momentum p, equals a constant. By picking your units carefully you can make this constant be the Planck constant h.

The point is that the more narrowly position is constrained to be, the broader is the constraint on momentum, and vice versa. In the limit where you know one of them exactly, as you know the momentum of the photons in a single-frequency beam of light, then the position of those photons becomes completely uncertain, you can't pick out one or the other in order; any of them could be anywhere along the beam.

9. Aug 27, 2006

Question. Why not also quantize the "position" to an operator--why just "momentum" ? What mathematics results if both position and momentum are simultaneously quantized ?

10. Aug 27, 2006

### Kurdt

Staff Emeritus
General relativity for me would fit the bill. Like quantum mechanics the leap in imagination for the concepts of general relativity was rather large. Also being a theory which makes use of tensors the scope of the field equation is enormous and also lends to an elegance that is beyond most other theories I've studied.

11. Aug 27, 2006

### Loren Booda

The displacement is also assigned an operator, usually the Kronecker delta function.

12. Aug 27, 2006

Thanks. So how does this knowledge modify this ? : $$\Delta x \Delta p = ih$$

13. Aug 27, 2006

### Schrodinger's Dog

Thanks that's a very complete answer and easy to understand, although I'm glad I just finished studying a statistics block or I may have had more questions.

14. Aug 27, 2006

### K.J.Healey

idk, maybe more accurate to say something:

$$\Delta x \Delta p > h/(2 \pi)$$

if I did that LaTeX right...

15. Aug 27, 2006

### Gokul43201

Staff Emeritus
I'm sure I do not understand. Could you explain?

16. Aug 27, 2006

### Loren Booda

Actually, the position operator in quantum mechanics is usually given by a Dirac delta function. It may be described 1.) as a function equal to zero except for one neighborhood of the domain approaching a singularity, with non-zero range, and 2.) so that the integral of its range over all its domain equals 1.

Take a sinusoidal wavefunction (technically, slowly tapering off as it approaches infinity, which allows that non-zero position probability can be measured at least finitely along its infinite extent, and normalization can be defined). Once measurement is made, the operator applies and the wavefunction "collapses," and a definite probability of 1 applies to the point of measurement.

Mathematically, the Dirac delta function may choose a discrete value of the variable (position) under consideration when integrating its probability over all space, effectively assigning certainty to the point of measurement. Likewise, position's complement, momentum, introduces an operator (the Fourier transform of the Dirac delta) which under integration yields an infinite range. One might say that a discrete frequency corresponds to a totally uncertain (infinite) wavetrain.