## my musings on tachyons.

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eSOANEM
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### my musings on tachyons.

I was bored on a school trip last week and was musing on tachyons when I came up with this idea about how they clearly violate all laws of reasonability.

Now I am aware that it is commonly stated that a tachyons moves backwards in time, why is this? Because by my calculations (see below) the movement through the temporal dimension should be imaginary rather than negative. If anyone could help me check this over, it would be much appreciated.

_______________________________________________________________________

In a spacetime 3d volume with 2 dimensions of space and 1 time, relativity tells us that the total magnitude of an objects motion is C (the speed of light). (I believe, hence the fact that movement through time slows down as you accelerate, verification of this would also be appreciated)

Using Pythagoras' theorem wear “a” is the distance travelled through space, “b” the distance travelled through time and “c” the overall magnitude in the spacetime volume we get the following:

a2+b2=c2

substituting in “a” as C+x (x positive real) and “c” as C we get the following:

(C+x)2+b2=C2

C2+2cx+x2+b2=C2

2Cx+x2+b2=0

as x and C are both positive then so must x2 and 2Cx be.

This implies that b2<0

This implies that b is imaginary not merely negative.

I cannot see any mistakes but then I cannot proof read my essays, html, maths or circuits well so if someone else could check this over that would be good.

_______________________________________________________________________

Of course, if the movement through time is imaginary then wouldn't that make the line impossible to plot without introducing another fourth imaginary axis to our spacetime volume to create a 4d complex spacetime volume.
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doogly
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### Re: my musings on tachyons.

What do you mean by 'magnitude of motion?' Length of the 4-velocity? (or three velocity as you would have it here)?
http://en.wikipedia.org/wiki/Four-velocity
If you want the length of any such vector, (a,b,c), it is going to be -a^2+b^2+c^2, or a^2-b^2-c^2, depending on your convention for the metric. But, what you definitely will have is a different sign for the time component than for the spacial components. With this sign change included you will not have imaginary entries for any component.
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sikyon
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### Re: my musings on tachyons.

eSOANEM wrote:I was bored on a school trip last week and was musing on tachyons when I came up with this idea about how they clearly violate all laws of reasonability.

Now I am aware that it is commonly stated that a tachyons moves backwards in time, why is this? Because by my calculations (see below) the movement through the temporal dimension should be imaginary rather than negative. If anyone could help me check this over, it would be much appreciated.

_______________________________________________________________________

In a spacetime 3d volume with 2 dimensions of space and 1 time, relativity tells us that the total magnitude of an objects motion is C (the speed of light). (I believe, hence the fact that movement through time slows down as you accelerate, verification of this would also be appreciated)

Using Pythagoras' theorem wear “a” is the distance travelled through space, “b” the distance travelled through time and “c” the overall magnitude in the spacetime volume we get the following:

a2+b2=c2

substituting in “a” as C+x (x positive real) and “c” as C we get the following:

(C+x)2+b2=C2

C2+2cx+x2+b2=C2

2Cx+x2+b2=0

as x and C are both positive then so must x2 and 2Cx be.

This implies that b2<0

This implies that b is imaginary not merely negative.

I cannot see any mistakes but then I cannot proof read my essays, html, maths or circuits well so if someone else could check this over that would be good.

What's wrong with imaginary space-time?

seladore
Posts: 586
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Location: Tumbolia

### Re: my musings on tachyons.

You can't treat a time-like dimension in the same way as a space-like one. So, you are going wrong when you create your 3D volume, and do normal Pythagorean calculations on it.

You need to use 4-space (I'm not sure why you only gave yourself two spatial dimensions), so there are three space-like dimensions and one time-like dimension.

The distance between two points in this 4-space, which we'll call ds, is

[math]ds^2 = -(dx^0)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2[/math]

Where [imath](x^0, x^1, x^2, x^3) = (ct, x, y, z)[/imath] in this example. So [imath]x^0[/imath] is your time-like dimension (which is ct, rather than just 't', so it is a length measurement).

Hope this helps.

I believe, hence the fact that movement through time slows down as you accelerate, verification of this would also be appreciated

You appear to do so to an external observer. To yourself, everything seems normal.

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