- #1

I couldn't get anyone to comment on the theory itself.

I found it online and it seems to make sense to me. I can't find any errors in the math or the reasoning.

The theory is available at http://members.triton.net/daveb

Please comment.

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- Thread starter etaylor
- Start date

- #1

I couldn't get anyone to comment on the theory itself.

I found it online and it seems to make sense to me. I can't find any errors in the math or the reasoning.

The theory is available at http://members.triton.net/daveb

Please comment.

- #2

Jonathan

- 365

- 0

I don't have the time to read the whole theory, but I read the first few pages and the diagram doesn't make sense to me. If I understand, the diagram is of the Feynman type, 2d, one space and one time. But space and time as shown by the diagram are not perpendicular. I think it should look like this:

E1(T)...E0(vT)

p3......p2

__vT__

\........|

.\.......|

..\......|

...\cT'.|

....\....|cT

.....\...|

......\..|

.......\.|

........\|

........E1,E0(0)

........p1

(ignore the dots)

Keep in mind that this is just what came to mind, if the diagram is wrong the math is probably wrong, but I probably misunderstood something. My whole problem with his diagram is that if the time axis is vertical (it is) and we draw objects E0 and E1 at time T, it makes sense that the line connecting p2 and 3 would be horizontal (perpendicular the the space axis), while his clearly isn't.

E1(T)...E0(vT)

p3......p2

__vT__

\........|

.\.......|

..\......|

...\cT'.|

....\....|cT

.....\...|

......\..|

.......\.|

........\|

........E1,E0(0)

........p1

(ignore the dots)

Keep in mind that this is just what came to mind, if the diagram is wrong the math is probably wrong, but I probably misunderstood something. My whole problem with his diagram is that if the time axis is vertical (it is) and we draw objects E0 and E1 at time T, it makes sense that the line connecting p2 and 3 would be horizontal (perpendicular the the space axis), while his clearly isn't.

Last edited:

- #3

The title of the paper say "non orthogonal" lorentzian geometry.

There is no means to measure the direction of time relative to space.

It has always been assumed to be perpendicular.

This theory assumes it is not, at least not for moving objects.

The math works and it seems to unify SR with the Dirac matrices.

I have read the whole paper 3 times and I am ready to put my money on this one.

I like it.

- #4

The title of the paper say "non orthogonal" lorentzian geometry.

There is no means to measure the direction of time relative to space.

It has always been assumed to be perpendicular.

This theory assumes it is not, at least not for moving objects.

The math works and it seems to unify SR with the Dirac matrices.

I have read the whole paper 3 times and I am ready to put my money on this one.

I like it.

- #5

jcsd

Science Advisor

Gold Member

- 2,101

- 12

- #6

Before Einstein published the special theory of relativity, the Lorentz transforms were well known. Einstein did not add any new equations (in that publications), he merely gave meaning to the Lorentz transforms.

Diracs matrices are well known and work well in relativistic quantum problems, however no one, including Dirac, has ever been able to give any physical meaning to them. Dirac himsself admitted that many times. He found then basically by mathematical trial and error very much like Lorentz did with his transforms.

What Barwacz has done is not only give them meaning, but also give a new tool to science (a non orthogonal representation of space tiime).

- #7

Jonathan

- 365

- 0

- #8

I can't imagine any method by which we could measure the direction of time relative to space. But if time is change, then it only makes sense that motion (which is a form of change) is the result of projecting a portion of time on to space.

As the Barwacz paper shows, doing so gives simple meaning to time dilation. Also, since the rate of time is fixed at a maximun, it would also clearly explain why nothing can travel faster than light. There simply isn't any more time to project on space.

The more I think about this paper the more sense it makes.

Does anyone else agree?

- #9

Jonathan

- 365

- 0

- #10

Can I say something here?

I don't know what your background is, and I haven't read the article you are refering to yet, but I can tell you this. It doesn't make any sense to ask whether time is perpendicular to space or not. Imagine a square and its diagonals. They are perpendicular to one another, but if you move to a velocity v close to the speed of light, c, the square will appear to shrink on your direction of movement (along one of the square's sides). So your square will have become a rectangle, and a rectangle's diagonales are not perpendicular to one another.

- #11

Brad_Ad23

- 502

- 1

For example, consider the space formed by the vectors

a=(1,0,0,0), b=(0,1,0,0), c=(0,0,1,0), t=(0,0,0,1)

To find out these vectors are orthonormal (perpendicular) we must take the inner product space.

<a|b> = Σ

Since we have 4 vectors, we wish to check if the vectors are all mutually perpindicular or if they are not.

<a|b|c|d> = Σ

In other words the sum of the products of the coefficients of the vectors of the i

So we have 1*0*0*0 + 0*1*0*0 + 0*0*1*0 + 0*0*0*1 = 0. All four vectors have an orthonormal inner product space. Hence we have 4 degrees of freedom, and four dimensions.

However, in quantum mechanics, time is regarded more as a parameter since it lacks a degree of freedom. This can be accounted for by saying that it is perpindicular, but on another space, an imaginary space, and is hence why one often sees the time dimension as a negative quantity when squared, or with an i next to it.

ex.

ds

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